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Why is an A-U bond less stable than an A-T bond?

Why is an A-U bond less stable than an A-T bond?


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I have encountered the following fact many times, but have not yet encountered a possible explanation for it. Will you please help me understand the molecular mechanism by which the bond between Adenine and Thymine is more stable than the bond between Adenine and Uracil?

Thanks in advance


I was not previously aware of this assertion, but on doing a little searching found it discussed in the following paper (I'm not sure whether it is open access or not).

Kamya PR and Muchall HM (2011) "Revisiting the effects of sequence and structure on the hydrogen bonding and π-stacking interactions in nucleic acids" J Phys Chem A. vol 115 pp. 12800-8

From this it emerges that the claim relates to the effect of substituting U for T in DNA, and the paper references previous claims for a stronger AT hydrogen bond. It also references other studies which did not find any difference, so the question was/is contentious. All studies were made on particular stretches of DNA duplex, and the authors' own studies were on a wider range of duplex structures from the protein data bank and (apparently) used more sophisticated analysis. To quote, they concluded:

"H-bonding in AT can be stronger or weaker than that in AU base pairs, depending on the particular choice of nucleic acids used for comparison"

They explain this in terms of the influence of neighbouring bases in the duplex on base stacking, which, depending on the bases (i.e. the sequence), can be enhanced by interactions with the methyl group of T. (There may also be other effects caused by distortion of the base-pairs from planarity in particular cases.) So it would appear that AT base-pairs are not intrinsically stronger than AU base-pairs, but can be stronger in certain three-dimensional contexts.

As a generalization, the premise of the question would therefore appear to be false.


Difference Between Thymine and Uracil

Thymine and Uracil are the two nucleotide bases which are found in the DNA and RNA respectively. The main difference between thymine and uracil is due to the property of “Occurrence”. The occurrence of both thymine and uracil is the property that distinguishes these two, as thymine is a pyrimidine nitrogenous base found in deoxyribonucleic acid (DNA). In contrast, Uracil is a pyrimidine nitrogenous base found in ribonucleic acid (RNA).

A deoxyribonucleic acid possesses cytosine and thymine as the pyrimidine bases, while a ribonucleic acid contains cytosine and uracil as the pyrimidine bases. You will get to know the key differences and similarities between the thymine and uracil nitrogenous base along with the comparison chart. Besides, we will also discuss the definition, structure, and important facts about both thymine and uracil.

Content: Thymine Vs Uracil

Comparison Chart

PropertiesThymineUracil
UracilIt is the pyrimidine base of the DNAIt is the pyrimidine base of the RNA
OccurrenceFound only in DNAFound only in RNA
Presence of keto groupTwo keto groups present at C-2 and C-4 atom Two keto groups present at C-2 and C-4 atom
Presence of methyl groupOne methyl group present at C-5 atomNot present
Molecular formulaC5H6N2O2C4H4N2O2
Molar mass126.1133 g/mol112.0868 g/mol

Definition of Thymine

The nitrogenous base comes under the type of pyrimidine base, which appears as a single ring structure within the backbone of deoxyribonucleic acid where it complements pairs with the purine base adenine by two hydrogen bonds. A methyl group is linked to the 5′-C of the thymine’s aromatic ring structure. It was first reported by the two scientists named Albrecht Kossel and Albert Neumann. Thymine was named after its presence in the calve’s thymus gland. Some of the chemical properties are mentioned below:

  • Chemical formula: C5H6N2O2
  • Molar mass: 126.115 g/mol -1
  • Density: 1.223 g/cm -3
  • Melting point: 316-317 degrees Celsius

Definition of Uracil

The nitrogenous base comes under the type of pyrimidine base, which appears as a single ring structure within the backbone of ribonucleic acid where it complements pairs with the purine base adenine by two hydrogen bonds. A hydrogen atom is linked to the 5′-C of the uracil’s aromatic ring structure. It was first reported by a scientist named Alberto Ascoli. Uracil was first derived by the hydrolysis of yeast nuclein. Some of the chemical properties are mentioned below:

  • Chemical formula: C4H4N2O2
  • Molar mass: 112.08 g/mol 1
  • Density: 1.32 g/cm 3
  • Melting point: 335 degrees Celsius

Structure of Thymine

Thymine is the pyrimidine base of the DNA, which contains two keto groups at C-2 and C-4 position and one methyl group at the C-5 position. It is denoted as T. It is synthesized by uracil methylation at the C-5 position of the pyrimidine ring, due to which thymine is also called 5-Methyl uracil. In DNA helix, the complementary pair of thymine is the purine base (adenine). The nitrogenous bases T and A pairs with each other by forming two hydrogen bonds.

Structure of Uracil

Uracil is the pyrimidine base of the RNA, which contains two keto groups at C-2 and C-4 position. It is denoted as U. In an RNA molecule, the complementary pair of uracil is the purine base (adenine). The nitrogenous bases U and A pairs with each other by the formation of two hydrogen bonds.

Important Facts

As we know that the combination of the nitrogenous bases and pentose sugar forms a nucleoside. The attachment of one or more phosphate groups to the nucleoside forms a nucleotide.


Introduction

Chemistry deals with the way in which subatomic particles bond together to form atoms. Chemistry also focuses on the way in which atoms bond together to form molecules. In the atomic structure, electrons surround the atomic nucleus in regions called orbitals. Each orbital shell can hold a certain number of electrons. When the nearest orbital shell is full, new electrons start to gather in the next orbital shell out from the nucleus, and continue until that shell is also full. The collection of electrons continues in ever widening orbital shells as larger atoms have more electrons than smaller atoms. When two atoms bond to form a molecule, their electrons bond them together by mixing into openings in each others' orbital shells. As with the collection of electrons by the atom, the formation of bonds by the molecule starts at the nearest available orbital shell opening and expand outward.


30 ATP: Adenosine Triphosphate

By the end of this section, you will be able to do the following:

  • Explain ATP’s role as the cellular energy currency
  • Describe how energy releases through ATP hydrolysis

Even exergonic, energy-releasing reactions require a small amount of activation energy in order to proceed. However, consider endergonic reactions, which require much more energy input, because their products have more free energy than their reactants. Within the cell, from where does energy to power such reactions come? The answer lies with an energy-supplying molecule scientists call adenosine triphosphate , or ATP . This is a small, relatively simple molecule ((Figure)), but within some of its bonds, it contains the potential for a quick burst of energy that can be harnessed to perform cellular work. Think of this molecule as the cells’ primary energy currency in much the same way that money is the currency that people exchange for things they need. ATP powers the majority of energy-requiring cellular reactions.

As its name suggests, adenosine triphosphate is comprised of adenosine bound to three phosphate groups ((Figure)). Adenosine is a nucleoside consisting of the nitrogenous base adenine and a five-carbon sugar, ribose. The three phosphate groups, in order of closest to furthest from the ribose sugar, are alpha, beta, and gamma. Together, these chemical groups constitute an energy powerhouse. However, not all bonds within this molecule exist in a particularly high-energy state. Both bonds that link the phosphates are equally high-energy bonds ( phosphoanhydride bonds ) that, when broken, release sufficient energy to power a variety of cellular reactions and processes. These high-energy bonds are the bonds between the second and third (or beta and gamma) phosphate groups and between the first and second phosphate groups. These bonds are “high-energy” because the products of such bond breaking—adenosine diphosphate (ADP) and one inorganic phosphate group (Pi)—have considerably lower free energy than the reactants: ATP and a water molecule. Because this reaction takes place using a water molecule, it is a hydrolysis reaction. In other words, ATP hydrolyzes into ADP in the following reaction:

Like most chemical reactions, ATP to ADP hydrolysis is reversible. The reverse reaction regenerates ATP from ADP + Pi. Cells rely on ATP regeneration just as people rely on regenerating spent money through some sort of income. Since ATP hydrolysis releases energy, ATP regeneration must require an input of free energy. This equation expresses ATP formation:

Two prominent questions remain with regard to using ATP as an energy source. Exactly how much free energy releases with ATP hydrolysis, and how does that free energy do cellular work? The calculated ∆G for the hydrolysis of one ATP mole into ADP and Pi is −7.3 kcal/mole (−30.5 kJ/mol). Since this calculation is true under standard conditions, one would expect a different value exists under cellular conditions. In fact, the ∆G for one ATP mole’s hydrolysis in a living cell is almost double the value at standard conditions: –14 kcal/mol (−57 kJ/mol).

ATP is a highly unstable molecule. Unless quickly used to perform work, ATP spontaneously dissociates into ADP + Pi, and the free energy released during this process is lost as heat. The second question we posed above discusses how ATP hydrolysis energy release performs work inside the cell. This depends on a strategy scientists call energy coupling. Cells couple the ATP hydrolysis’ exergonic reaction allowing them to proceed. One example of energy coupling using ATP involves a transmembrane ion pump that is extremely important for cellular function. This sodium-potassium pump (Na + /K + pump) drives sodium out of the cell and potassium into the cell ((Figure)). A large percentage of a cell’s ATP powers this pump, because cellular processes bring considerable sodium into the cell and potassium out of it. The pump works constantly to stabilize cellular concentrations of sodium and potassium. In order for the pump to turn one cycle (exporting three Na+ ions and importing two K + ions), one ATP molecule must hydrolyze. When ATP hydrolyzes, its gamma phosphate does not simply float away, but it actually transfers onto the pump protein. Scientists call this process of a phosphate group binding to a molecule phosphorylation. As with most ATP hydrolysis cases, a phosphate from ATP transfers onto another molecule. In a phosphorylated state, the Na + /K + pump has more free energy and is triggered to undergo a conformational change. This change allows it to release Na + to the cell’s outside. It then binds extracellular K + , which, through another conformational change, causes the phosphate to detach from the pump. This phosphate release triggers the K + to release to the cell’s inside. Essentially, the energy released from the ATP hydrolysis couples with the energy required to power the pump and transport Na + and K + ions. ATP performs cellular work using this basic form of energy coupling through phosphorylation.

One ATP molecule’s hydrolysis releases 7.3 kcal/mol of energy (∆G = −7.3 kcal/mol of energy). If it takes 2.1 kcal/mol of energy to move one Na + across the membrane (∆G = +2.1 kcal/mol of energy), how many sodium ions could one ATP molecule’s hydrolysis move?

Often during cellular metabolic reactions, such as nutrient synthesis and breakdown, certain molecules must alter slightly in their conformation to become substrates for the next step in the reaction series. One example is during the very first steps of cellular respiration, when a sugar glucose molecule breaks down in the process of glycolysis. In the first step, ATP is required to phosphorylze glucose, creating a high-energy but unstable intermediate. This phosphorylation reaction powers a conformational change that allows the phosphorylated glucose molecule to convert to the phosphorylated sugar fructose. Fructose is a necessary intermediate for glycolysis to move forward. Here, ATP hydrolysis’ exergonic reaction couples with the endergonic reaction of converting glucose into a phosphorylated intermediate in the pathway. Once again, the energy released by breaking a phosphate bond within ATP was used for phosphorylyzing another molecule, creating an unstable intermediate and powering an important conformational change.

See an interactive animation of the ATP-producing glycolysis process at this site.

Section Summary

ATP is the primary energy-supplying molecule for living cells. ATP is comprised of a nucleotide, a five-carbon sugar, and three phosphate groups. The bonds that connect the phosphates (phosphoanhydride bonds) have high-energy content. The energy released from ATP hydrolysis into ADP + Pi performs cellular work. Cells use ATP to perform work by coupling ATP hydrolysis’ exergonic reaction with endergonic reactions. ATP donates its phosphate group to another molecule via phosphorylation. The phosphorylated molecule is at a higher-energy state and is less stable than its unphosphorylated form, and this added energy from phosphate allows the molecule to undergo its endergonic reaction.

Visual Connection Questions

(Figure) One ATP molecule’s hydrolysis releases 7.3 kcal/mol of energy (∆G = −7.3 kcal/mol of energy). If it takes 2.1 kcal/mol of energy to move one Na + across the membrane (∆G = +2.1 kcal/mol of energy), how many sodium ions could one ATP molecule’s hydrolysis move?

(Figure) Three sodium ions could be moved by the hydrolysis of one ATP molecule. The ∆G of the coupled reaction must be negative. Movement of three sodium ions across the membrane will take 6.3 kcal of energy (2.1 kcal × 3 Na + ions = 6.3 kcal). Hydrolysis of ATP provides 7.3 kcal of energy, more than enough to power this reaction. Movement of four sodium ions across the membrane, however, would require 8.4 kcal of energy, more than one ATP molecule can provide.

Review Questions

The energy released by the hydrolysis of ATP is____

  1. primarily stored between the alpha and beta phosphates
  2. equal to −57 kcal/mol
  3. harnessed as heat energy by the cell to perform work
  4. providing energy to coupled reactions

Which of the following molecules is likely to have the most potential energy?

Critical Thinking Questions

Do you think that the EA for ATP hydrolysis is relatively low or high? Explain your reasoning.

The activation energy for hydrolysis is very low. Not only is ATP hydrolysis an exergonic process with a large −∆G, but ATP is also a very unstable molecule that rapidly breaks down into ADP + Pi if not utilized quickly. This suggests a very low EA since it hydrolyzes so quickly.

Glossary


Better Than the Bank?

The interest rates on bonds are typically greater than the deposit rates paid by banks on savings accounts or CD. As a result, if you are saving and you don’t need the money in the short term (in a year or less), bonds will give you a relatively better return without posing too much risk.

College savings are a good example of funds you may want to increase through investment, while also protecting them from risk. Parking your money in the bank is a start, but it’s not going to give you any return. With bonds, aspiring college students (or their parents) can predict their investment earnings and determine the amount they’ll have to contribute to accumulating their tuition nest egg by the time college starts.

Bonds do have credit risk and are not FDIC insured as are bank deposit products.   Therefore, you do have some risk that the bond issuer will go bankrupt or default on their loan obligations to bondholders. If they do, there is no government guarantee that you'll get any of your money back.


10 Answers 10

No, what it says is “In half a year, we’ll give whoever holds this bond $10,000 for which you pay us $9,750 now”.

This is equivalent to an annual interest rate of about 5% (the example is showing a yield way more than currently available). Plus you can sell the bond to another person in the meantime.

Treasury Bills (T-bills) does seem like an oddball but it might work for some folks.

I'm going to address it both your questions individually as interest and liquidity.

Looking at the Department of Treasury's site for rates (https://www.treasury.gov/resource-center/data-chart-center/interest-rates/Pages/TextView.aspx?data=billrates, Jan 8, 2019) I see the range is 2.40-2.60% depending on the timeframe (4 weeks - 52 weeks).

This appears mostly comparable to the top savings rates (2.00-2.45%, viewed Jan 8, 2019) found on Bankrate: https://www.bankrate.com/banking/savings/rates/

From an interest perspective, it's essentially the same. What needs to get taken into account are the amounts involved and any special hoops to jump through to avoid fees.

Some of the special savings rates found on Bankrate require a minimum balance. The top rate (2.45%) requires $25,000 balance while the second highest rate (2.39%) requires only $1. Each financial institution (FI) may have their own requirements to avoid any maintenance fees: electronic statements, use debit card X/mo, direct deposit, etc.

For amounts under $250,000, your money is insured by the FDIC (banks) or NCUA (credit unions) if your FI fails. For funds above that, you'll need to either open additional accounts or have them at other FIs to keep your money safe.

T-bills can be purchased in $100 increments, so there is no minimum balance requirement other than the purchase itself. It's also guaranteed by the US Government, so it's considered a risk-free investment (https://www.investopedia.com/ask/answers/013015/how-are-treasury-bills-taxed.asp).

If you have more than $250,000 that you want saved/invested "risk-free", then T-bills could be an option for this, outside the normal channels.

So keeping cash under the mattress has the advantage of being extremely liquid: you can take it out whenever you need it. It has the disadvantages of being insecure and losing value due to inflation (and maybe logistics if you have a very large sum of money).

Savings accounts are also very liquid with one catch: you are limited to six withdrawals per month per Reg D (https://www.nerdwallet.com/blog/banking/how-regulation-d-affects-your-savings-withdrawals/). Transactions in person or ATM don't count in this limit. Your money is kept safe and insured up to the $250,000 limit.

T-bills can be purchased in increments of 4-, 8-, 13-, 26-, and 52-weeks. You won't be able to access your money during that time, but you also won't lose it either unless the US Government defaults. It's the same concept as Certificates of Deposit (CDs).

Hopefully this helps answer why someone might choose one over the other.

Other answers explain why it would be interesting to buy Treasury Bills, I am going to explain the "discount" concept.

The point here is the mechanism of how those TB are sold to the primary market (big investors, banks, and other entities that will sell the TBs to small investors).

The Treasury is issuing promissory notes of paying a quantity (let's say $1,000) sometime in the future and auctioning them.

If demand is high (instability is high, or other investments give low ROIs) then the primary market will offer more for those notes, so the Treasury gets paid more (let's say $970) the interest for the buyer will be $30.

If demand is low (there are other interesting investments elsewhere) the price drops and and the Treasury will get pay less (let's say $920) the interest for the buyer will be $80.

The difference between the nominal value of the bill and its purchase price is its "discount" (you could read it as "how much the Treasury discounts the bill in order to sell it").

Of course, for the small investor that distinction is meaningless in relation to an usual "bonus": either he is buying in the secondary market (from the primary buyers) and has to accept a discount that has already been set, or even if he can buy in the primary market 1 the volume that he is going to buy will not move the price that will be set in the auction 2 .

1 I am not sure about the USA, but in some countries it is possible.

2 And there are risks with this system, too:

If the offer is too low the investor will not get any bills.

In the USA (thanks @dave_thompson_085 for the info) and probably other countries, the price is set by accepting bids from higher price to lower price until the bills offered are sold, and the final price for all of the buyers is that of the accepted offer with a lower value. That means that the investors do not know which is the actual purchase price they have commited themselves to (although they do know that at worst it is as high as their bid).

For average individuals there isn't much point to this apart from the small interest.

For large amounts of money, there is the problem that bank insurance only applies up to a certain dollar value - currently $250k. Beyond that you're exposed to the risk of the bank defaulting.

Keeping large amounts of physical cash is even more of a risk, against threats internal and external. And it costs money (staff and security time) to move it in and out. Cash handling is expensive even in businesses that do it all the time like supermarkets and casinos.

If you have a lot of money, or if you are the bank, then bonds are much simpler and less risky. They're also available in conveniently large denominations.

That non-specific example illustrates about 5% annually which is pretty good and about double the current actual market for a 6-month treasury.

Treasuries are marketable securities so you can sell it whenever you feel for whatever the market rate is at the time.

For the last few years, the interest rate of Japanese government bond is negative. That is, you give 100 to them, and they pay you back 99 in 5 years.

The top concern other than earning interest is safety. If you put your money under the bed, it might be stolen or burnt down. Besides, large institutions simply can't have billions in bank notes stored in their office. If you put your money in a commercial bank to earn small interest, the bank might fall and you never get your money back.

The treasury bill or other government bonds are backed by the government, who has the authority to print money, so they won't go broke and you always get your money back. (The exception is euro area where governments can't freely print money. Another exception is Russia in the late 90s when they decided that defaulting on their own gov bond was better to the economy than printing quadrillion rubles. Even the US treasury carries the risk of technical default once in a while.)

The advantages of the discount auction become more apparent when you consider the number banks are dealing with. You might look at putting $10,000 in a 6 month T-bill, but a bank is going to be looking at numbers in the billions.

At a 2% discount, you hand Treasury $9,800 and that at the end of the term (let's say 1 year to make it simple) you get $10,000 back. That means you are able to spend that discounted $200, invest it elsewhere, or lend it to someone else.

A bank putting $2 billion in the same security will get to use $40 million for other things. It's a subtle but important difference.

Another point to consider is taxes. Interest paid by the US Federal Government is exempt from State taxes and therefore has a overall better yield than an equivalent interest rate paid by the banks.

It makes sense to use them in a portfolio strategy. Think of a portfolio as a recipe, and t-bills are an ingredient.

When you make a portfolio you are looking to maximize return at a certain level of risk, minimize risk at a certain level of return, or some optimal combination of the two, related to the clients risk profile (I.e. if you are 25 you should take some risks to maximize returns, if things go bad you have time to make it up, if you are 65 you can’t take those risks, therefore your potential return is also diminished)

Portfolios are recipes that generate a less uncertain (but lower) result when they use two assets that are negatively correlated. The classic example is airlines and oil companies: When oil gets cheap airlines make bank, and oil companies hurt. When oil gets expensive it’s the other way. Buy stocks on BOTH sectors and you are mostly immune to variations in the price of oil.

So what is negatively correlated with T-Bills? Think of the value of money itself. When the economy overheats there is more demand than there are goods and services available for purchase, so prices go up. That is inflation, it means money is worth less than before, and your T-Bills can actually lose value in terms of their real purchase power. So who gains in that scenario? The people making the goods and services that are in such high demand that the prices keep going up. In other words, stocks in companies like airlines and oil producers.

Now think of the opposite scenario: bankers get jealous of the bonuses hedge fund guys were getting, so they lobby for financial deregulation, doing away with the protections that kept the system stable. Their reckless gambling wrecks the economy, people lose their jobs, the housing market crashes and people stop buying stuff. Now there is cheap stuff to buy everywhere, because nobody has money to buy anything. In that scenario the value of the money invested in T-bills went UP, you can buy more stuff with the same money. Stocks in the other hand crashed, because nobody has the money to buy stuff. (Except for the bankers. It wasn’t their own money they were gambling with after all, but there aren’t enough of them). In that crazy scenario (aka 2008) your portfolio still worked, one cancelled the other.


12 Bond Mutual Funds and ETFs to Buy for Protection

When the stock market took a beating this spring, nervous investors looked to bond mutual funds and exchange-traded funds (ETFs) for protection and sanity. After all, fixed income typically provides regular cash and lower volatility when markets hit turbulence.

And the markets absolutely hit turbulence. For instance, between Feb. 19 and March 10, not only did the S&P 500 experience a historically rapid loss of 14.8% – it experienced a dramatic rise in volatility, too, hitting its highest level on that front since 2011, says Jodie Gunzberg, chief investment strategist at New York-based Graystone Consulting, a Morgan Stanley business. The index's losses and volatility escalated even more through the March 23 lows.

However, bonds offer ballast – "not only downside protection but also moderate upside potential as investors tend to seek out the safety of U.S. government and investment-grade corporate bonds amid stock market uncertainty" – says Todd Rosenbluth, senior director of ETF and mutual fund research at CFRA, a New York-based investment research company.

Bond prices often are uncorrelated to equities. Stocks typically do well in periods of economic growth, whereas bonds typically do well in periods of declining economic activity, Gunzberg says. The Federal Reserve has also thrown in its support, buying up corporate bonds and even bond ETFs over the past couple months, in turn driving up private purchases of debt.

Indeed, bond funds have done extremely well in 2020. Of course, yields have thinned out and their room for upside has shrunk as a result. Nonetheless, investors still can find stability and some measure of income in these products.

Here are 12 bond mutual funds and bond ETFs to buy. These funds offer diversified portfolios of hundreds if not thousands of bonds, and most primarily rely on debt such as Treasuries and other investment-grade bonds. Just remember: This is an unprecedented environment, and even the bond market is acting unusually in some areas, so be especially mindful of your own risk tolerance.

Returns and data are as of July 21, unless otherwise noted. For mutual funds, returns and data are gathered for the share class with the lowest required minimum initial investment – typically the Investor share class or A share class. If you use an investment adviser or online brokerage, you may be able to buy lower-cost share classes of some of these funds. Yields are SEC yields, which reflect the interest earned after deducting fund expenses for the most recent 30-day period and are a standard measure for bond and preferred-stock funds.


Chymosin, Pepsins and Other Aspartyl Proteinases: Structures, Functions, Catalytic Mechanism and Milk-Clotting Properties

Therese Uniacke-Lowe , Patrick F. Fox , in Cheese (Fourth Edition) , 2017

Chymosin

As far as is known, the milk of all species coagulates in the neonatal stomach to increase the efficiency of digestion and to spread the ingestion of milk more evenly. The neonatal stomach of most, but not all, species secretes a special gastric aspartyl proteinase, chymosin , to induce coagulation. Chymosin (formerly known as “rennin” and not to be confused with renin, EC 3.4.23.4, an enzyme in the kidney) is commercially important for cheesemaking, a process which dates back to ∼6000–7000 BC when widespread domestication of sheep and goats occurred throughout the Fertile Crescent although initially animals were probably raised for meat and not milk production as adult lactose intolerance was almost universal ( Kindstedt, 2012 ). By 6500 BC substantial shifts from meat to milk production occurred in western Anatolia which coincided with the discovery of pyrotechnology and the production of ceramics. Analysis of organic residues from pottery shards from the archaeological strata of 6500–6000 BC showed the presence of processed dairy products, probably cheese and butter or ghee, in ceramic pots ( Kindstedt, 2012 ). Milk stored in ceramic pots would have fermented quickly and coagulated spontaneously rendering the cheese or curd digestible by lactose-intolerant adults. Legend has it that the first use of rennet can be traced back to a traveler who, transported milk inside a vessel made from a dried calf or lamb stomach. In transit, the rennet from the stomach transformed the liquid milk into curds and whey, a process that made cheese what it is today: a nutrient-rich food that transports well and can be kept for years ( Dunn, 2010 Tamime, 1993 ). This scenario seems unlikely as it presupposes that the milk-drinking nomad was lactose-tolerant ( Kindstedt, 2012 ). A more plausible explanation for the origin of cheesemaking using rennet is that Neolithic herders would have been well aware that the stomachs of kids, lambs, and calves, which were slaughtered or died naturally, were filled with coagulated milk thus providing ample opportunity to establish an association between the stomach, milk coagulation, and cheese curd ( Kindstedt, 2012 ).

The active “ingredients’ in this process were identified as the proteolytic enzymes chymosin and pepsin ( Foltmann, 1966 ), referred to as “rennet” ( Fruton, 2002 ). Chymosin is produced in utero in the abomasal mucosa of foetal mammals ( Foltmann, 1970 ), whereas pepsin (EC 3.4.23.1) is predominant in adult mammalian gastric secretions ( Rampili et al., 2005 ). It has been reported that the secretion of chymosin in the calf abomasum is stimulated by the presence of bovine milk in the calf’s stomach and that the casein fraction is responsible for this ( Garnot et al., 1977 ). Replacement of casein by fish or soy proteins in a milk substitute did not influence the pepsin content ( Garnot et al., 1974 ) or rate of secretion of pepsin but significantly affected both the amount and rate of secretion of chymosin in the calf’s stomach ( Garnot et al., 1977 ). These workers hypothesized that casein degradation products, such as phosphopeptides or caseinomacropeptide (CMP), may be responsible for increased chymosin activity in the calf stomach. Milk-feeding calves rather than cereal- or hay-feeding is reported to prolong prochymosin production in the calf’s abomasal mucosa to a certain extent ( Andrén, 1992 ).

Chymosin represents >90% of the milk-clotting activity of good quality veal rennet, the remaining activity being due to pepsin. As the animal ages, especially when fed solid food, the secretion of chymosin declines while that of pepsin increases. Like many other animal proteinases, chymosin is secreted as its zymogen, prochymosin, which is autocatalytically activated on acidification to pH 2–4 by removal of a 44-residue peptide from the N-terminal of the zymogen.

Foltmann and Pedersen (1977) proposed a classification of pepsins and pepsin-like enzymes based on their optimum pH on specific substrates. Martin et al. (1980) reported an improved classification system for these enzymes based on kcat/KM ratio obtained when a synthetic hexapeptide, Leu-Ser-Phe(NO2)-Nle-Ala-Leu-OMe, was used as substrate. Class I enzymes were those with a kcat/KM > 100 mM/s and included porcine and bovine pepsins and bovine gastricin, while Class II enzymes had a kcat/Km < 100 mM/s and included chymosin and proteinases from Rhizomucor miehei and Rhizomucor pusillus.

Chymosin has high milk-clotting activity but low general proteolytic activity and has so evolved to avoid proteolysis and inactivation of the immunoglobulins in colostrum ( Foltmann and Axelsen, 1980 Foltmann, 1981a ). Exceptions are some primates, including humans which, apparently, secrete the more proteolytic enzyme, pepsin, from birth the neonate of humans and other primates receives Igs in utero and hence dietary Ig from colostrum is less important than in these species. The rabbit and guinea pig also transfer Ig in utero but ruminants, the horse and pig transfer Ig via colostrum while the dog, rat, and mouse transfer Ig both in utero and via colostrum ( Hurley and Theil, 2013 Larson, 1992 Pentsuk and van der Laan, 2009 ). Neonatal kittens rely on colostrum for passive transfer of immunoglobulins ( Casal et al., 1996 Claus et al., 2006 Harding et al., 1961 Levy et al., 2001 ). The ratio between the general proteolytic activity and milk-clotting activity for cat chymosin is about 10 fold less than that of cat pepsin, supporting the hypothesis that chymosin is produced postnatally in these species to ensure that IgG uptake is not hindered ( Jensen et al., 1982 ).

Calf chymosin is well characterized at the molecular and enzymatic levels the extensive literature has been the subject of several reviews, including Chitpinityol and Crabbe (1998) Crabbe (2004) Foltmann (1966, 1970, 1971, 1981a, 1987, 1992, 1993) Jacob et al. (2011) Szecsi and Harboe (2013) and Yegin and Dekker (2013) . The enzyme was crystallized by Berridge (1945) Bunn et al. (1971) Ernstrom (1958) Foltmann (1958) and Hankinson (1943) free boundary electrophoresis showed that some early crystalline preparations were not homogeneous ( Ernstrom, 1958 ). Heterogeneity was not explained but may have been due to isoenzymes (see later large batches of commercial rennet were used) or to inactivated enzyme or other proteins. Calf chymosin is a single-chain polypeptide containing 323 amino acid residues with a molecular mass of 35,600 Da. Its primary structure has been established and a considerable amount of information is available on its secondary and tertiary structures. The secondary structure is 13% helical (9 helices, 44 residues) and 48% β-sheet (29 strands, 158 residues) ( Palmer et al., 2010 ). The protein contains three disulfides (Cys47-Cys52, Cys207-Cys211 and Cys250-Cys283) and a cis-proline (Pro25), which is conserved in mucorpepsin, endothiapepsin, and porcine pepsin (see Palmer et al., 2010 ). An illustration of the three-dimensional structure of chymosin is shown in Fig. 4.10 . As described earlier for aspartyl proteinases in general, the molecule exists as two domains separated by the active site cleft in which the two catalytically active aspartyl residues (Asp34 and Asp216) are located with their side chains oriented toward the cleft. The active cleft can accommodate ∼7 residues of the κ-casein sequence ( Fig. 4.11 ). The distance between the carboxyl oxygen of the two Asp residues is 3.1 Å. The carboxyl group of the Asp residues is connected via a complex network of hydrogen bonds and a number of water molecules in the immediate environment of the active site cleft ( Gilliland et al., 1990 Mantafounis and Pitts, 1990 ). In the tertiary structure, two adjacent threonine residues, Thr35 and Thr217, are hydrogen bonded to each other via the side chain oxygen and the main chain nitrogen, while their side chains are directed in the hydrophobic pocket to from the “fireman’s grip” ( Pearl and Blundell, 1984 see Fig. 4.6 ), that is, the side-chain oxygen of one Thr is bonded to the main chain nitrogen of the corresponding Thr, which provides additional structural stability to the catalytic site (see Chitpinityol and Crabbe, 1998 Wong, 1995 ). A water molecule is found between the catalytic Asp residues and is involved in the catalytic mechanism (see Section Catalytic Mechanism of Aspartyl Proteinases). Conserved water molecules (attached to residues 502 and 513 pepsin numbering) extend the fireman’s grip by making additional hydrogen bonds, and provide additional stability for the fireman’s grip ( Prasad and Suguna, 2002 ). The rigidity of the active-site geometry and of the fireman’s grip are essential for the activity of aspartyl proteinases and it has been reported that collapse of the fireman’s grip in proplasmepsin II makes the enzyme inactive as Asp residues are pulled away from each other due to reorientation of the enzyme domains ( Bernstein et al., 1999 Prasad and Suguna, 2002 ).

Figure 4.10 . The three-dimensional structure of chymosin (generated from PDB ID: 4CMS Newman et al., 1991 ).

Residues critical for catalytic activity are highlighted in yellow. Disulfide bridges are highlighted in red.

(Image from Yegin, S., Dekker, P., 2013. Progress in the field of aspartic proteinases in cheese manufacturing: structures, functions, catalytic mechanism, inhibition and engineering. Dairy Sci. Technol., 93, 565–594.)


Sandwalk

The four bases of DNA can exist in at least two tautomeric forms as shown below. Adenine and cytosine (which are cyclic amidines) can exist in either
amino or imino forms, and guanine, thymine, and uracil (which are cyclic amides) can exist in either lactam (keto) or lactim (enol) forms. The tautomeric forms of each base exist in equilibrium but the amino and lactam tautomers are more stable and therefore predominate under the conditions found inside most cells. The rings remain unsaturated and planar in each tautomer.


Fifty years ago it wasn't clear whether the amino or imino forms of the purines were stable under physiological conditions. (Or the lactam lactim forms.) As we will see, this uncertainty played a significant role in events leading up to the discovery of the structure of DNA.

We now know that all of the bases in the common nucleotides can participate in hydrogen bonding. The amino groups of adenine and cytosine are hydrogen donors, and the ring nitrogen atoms (N-1 in adenine and N-3 in cytosine) are hydrogen acceptors (see below). Cytosine also has a hydrogen acceptor group at C-2. Guanine, cytosine, and thymine can form three hydrogen bonds. In guanine, the group at C-6 is a hydrogen acceptor, and N-1 and the amino group at C-2 are hydrogen donors. In thymine, the groups at C-4 and C-2 are hydrogen acceptors, and N-3 is a hydrogen donor. (Only two of these sites, C-4 and N𔃁, are used to form base pairs in DNA.) The hydrogen-bonding patterns of bases have important consequences for the three-dimensional structure of nucleic acids.


Important Questions for CBSE Class 12 Chemistry – The p-Block Elements

PREVIOUS YEARS QUESTIONS

Very Short Answer Type Questions [1 Mark]

Question 1:
Write the formulae of any two oxoacids of phosphorus.
Answer:
H3P03 and H3PO4.

Question 2:
Write the formulae of any two oxoacids of chlorine.
Answer:
HClO3 and HCl04.

Question 3:
H3P03 disproportionates while H3P04 does not, why?
Answer:
It is because in H3P03, ‘P’ is in +3, intermediate oxidation state which can increase to +5 and decrease to -3, whereas in H3P04, ‘P’ is in highest oxidation state +5 which can only gain electrons, i.e. undergoes reduction only, acts as oxidising agent and cannot disproportionate.

Question 4:
Out of white phosphorus and red phosphorus, which one is more reactive and why?
Answer:
White phosphorus because it is monomeric and has low bond dissociation enthalpy due to angle of strain (bond angle 60°).

Question 5:
What is the basicity of H3P04?

Answer:
Three.

Question 6:
Write the formulae of any two oxoacids of sulphur.
Answer:
H2S04 and H2SO3.

Question 7:
Which allotrope of sulphur is thermally stable at room temperature?
Answer:
Rhombic sulphur.

Short Answer Type Questions [I] [2 Marks]

Question 8:
Write the structures of the following molecules: (i) H2SO3 (ii) XeOF4
Answer:

Question 9:
Write the structures of the following:
(i) H2S207 (ii) Xe03
Answer:

Question 10:
Write the structures of the following:
(i) N2O5 (ii) BrF3
Answer:

Short Answer Type Questions [II] [3 Marks]

Question 11:
Give reasons for the following:
(i) N2 is less reactive at room temperature.
(ii) H2Te is the strongest reducing agent amongst all the hydrides of group 16-elements.
(iii) Helium is used in diving apparatus as a diluent for oxygen.
Answer:
(i) It is due to presence of triple bond which has high bond dissociation enthalpy.
(ii)H2Te has longest bond length which has lowest bond dissociation enthalpy.
(iii) It is because helium is less soluble than N2 in blood and does not cause pain.

Question 12:
Give reasons for the following:
(i) NH3 has a higher boiling point than PH3.
(ii) H2Te is more acidic than H2S.
(iii) Chlorine water on standing loses its yellow colour.
Answer:
(i) NH3 is associated with inter molecular H-bonding, PH3 is not.
(ii) H2Te has lower bond dissociation enthalpy than H2S due to longer bond length.
(iii) Cl2+H2O —–> HCL+HOCL
If forms HCl and HOCl, both are colourless.

Question 13:
(a) Account for the following:
(i) Bond angle in NH4 is greater than that in NH3.
(ii)Reducing character decreases from S02 to TeO2.
(iii) HClO4 is a stronger acid than HCIO.
(b) Draw the structures of the following:
(i) H2S2O8
(ii) XeOF4.
Answer:
(a) (i) NH3 has lone pair of electron, so, bond angle is 107°, whereas NH +4 does not, therefore, bond angle is 109.5°.
(ii) It is due to stability of higher oxidation state which decreases due to inert x , effect.
(iii) It is because CIO4 – is more stable than CIO- due to more dispersal charge on four oxygen atoms.

Question 14:
(a) Which poisonous gas is evolved when white phosphorus is heated with Cone. NaOH solution? Write the chemical equation.
(b) Write the formula of first noble gas compound prepared by N. Bartlett. What inspired N. Bartlett to prepare this compound?
(c) Fluorine is a stronger oxidising agent than chlorine. Why?
(d)Write one use of chlorine gas.
(e)Complete the following equation:
CaF2 + H2S04 ———>
Answer:

(b)Xe+PtFg. The comparable ionisation enthalpy of 02 molecule (1175 KJ mol-1) and Xe (1170 KJ mol-1) inspired Neil Bartlett to prepare this compound.
(c) It is due to low bond dissociation enthalpy, higher hydration energy of F- and high electron gain enthalpy.
(d) It is used as bleaching agent and disinfectant.
(e) CaF2 + H2S04 ———–> CaS04 + 2HF.

Question 15:
(a) Account for the following:
(i) Bi(V) is stronger oxidizing agent than Sb(V).
(ii) H—O—I is a weaker acid than H—O—Cl.
(iii) Bond angle decreases from H2O to H2S.
(b)Draw the structures of the following:
(i) SF4
(ii) XeF2
Answer:
(a) (i) Bi3+ is more stable than Sb3+ due to inert pair effect. Bi5+ can gain 2
electrons to form Bi3+. That is why Bi5+ is stronger oxidising agent than Sb5+.
(ii) It is because ‘Cl’ is more electronegative than ‘I’.
(iii) It is because oxygen is more electronegative and smaller in size than sulphur.

Question 16:
(i) Why does PCl5 fume in moisture?
(ii) Write the name of the allotrope of sulphur which is stable at room temperature.
(iii) Chlorine water on standing loses its yellow colour. Why?
(iv) Write the disproportionation reaction of H3P03.
(v) Complete the following equation: F2 + H20 ———>
Answer:

Question 17:
(a) (i) Acidic character increases from HF to HI.
(ii) There is large difference between the melting and boiling points of oxygen and sulphur.
(iii) Nitrogen does not form pentahalide.
(b) Draw the structures of the following:
(i) CIF3 (ii) XeF4
Answer:
(a) (i) It is because bond dissociation energy decreases due to increase in bond length as atomic size of halogen increases from HF to HI.
(ii) Oxygen is diatomic gas having weak van der Waals’ forces, whereas sulphur is octaatomic (S8) solid, therefore, it has more van der Waals’ forces of attraction. Hence sulphur has higher melting and boiling points than oxygen.
(iii) It is because nitrogen does not have d-orbitals.

Question 18:
(i) Which allotrope of phosphorus is more reactive and why?
(ii) How the supersonic jet aeroplanes are responsible for the depletion of ozone layers?
(iii) F2 has lower bond dissociation enthalpy than Cl2. Why?
(iv) Which noble gas is used in filling balloons for meteorological observations?
(v) Complete the equation:
XeF2 + PF5 ——->
Answer:
(i) White phosphorus because it is monomeric and has low bond dissociation enthalpy due to angle of strain (bond angle 60°).
(ii) Supersonic jet aeroplanes release NO which is responsible for the depletion
of ozone layer. NO + 03 ——-> N02 + 02
(iii) It is due to more inter electronic repulsion between lone pair of electrons
(iv) Helium.
(v) XeF2 + PF5 ——-> [XeF]+ [PF]-

Question 19:
(a) Account for the following:
(i) Bond angle in NH4+ is higher than NH3.
(ii) H2S has lower boiling point than H20.
(iii) Reducing character decreases from S02 to TeO2.
(b) Draw the structure of the following
(i) H4P207 (pyrophosphoric acid) (ii) XeF2
Answer:
(a) (i) Refer Ans. to Q. 13 (a) (i).
(ii) H2S molecules are not associated with intermolecular H-bonding and have weak van der Waals’ forces of attraction, therefore, H2S has lower boiling point than H20 in which molecules are associated with intermolecular H-bonding.
(iii) Refer Ans. to Q. 13 (a) (ii).

Question 20:
(
a) Draw the structures of the following:
(i) XeF4 (ii) H2S207
(b) Account for the following:
(i) Iron on reaction with HCl forms FeCl2 and not FeCl3.
(ii) HC04 is a stronger acid than HCIO.
(iii) BiH3 is the strongest reducing agent amongst all the hydrides of group 15.
Answer:

(b ) (i) Fe reacts with HCl to form FeCl2 because HCl is not an oxidising agent.
Secondly, if any FeCl3 is formed, it will be reduced to FeCl2 by [H] [nascent hydrogen],
(ii) Refer Ans. to Q.13 (a) (iii)
(iii) BiH3 has lowest bond dissociation enthalpy due to longer bond length. Therefore, it acts as strongest reducing agent.

2014

Very Short Answer Type Questions [ 1 Mark ]

Question 21:
What is the basicity of H3P03?
Answer:
H3PO3 is dibasic acid. Its basicity is 2.

Question 22:
Why does N02 dimerise?
Answer:
It is because NO2 has unpaired (odd) electron, therefore, it is unstable and forms dimer to become stable.

Question 23:
Why does NH3 act as a Lewis base?
Answer:
It is because in NH3, there is lone pair of electrons on ‘N’ therefore, it acts as Lewis base.

Question 24:
Why is the single N—N bond weaker than the single P—P bond?
Answer:
There is more repulsion between lone pair of electrons on smaller ‘N’ atoms
in :N—N: bond due to which it is weaker than :P—P: bond.

Question 25:
Arrange the following in the increasing order of their basic character:
NH3, PH3, ASH3, SbH3, BiH3
Answer:
BiH3 < SbH3 < ASH3 < PH3 < NH3.

Short Answer Type Questions [I] [2 Marks]

Question 27:
Arrange the following in the order of property indicated against each set:
(i) HF, HCl, HBr, HI – increasing bond dissociation enthalpy.
(ii) H20, H2S, H2Se, H2Te – increasing acidic character
Answer:
(i) HI < HBr < HCl < HF is the increasing order of bond dissociation enthalpy.
(ii) H2O < H2S < H2Se < H2Te is the increasing order of acidic character.

Question 28:
Complete the following equations:
(i) P4 + H2O ———>
(ii) XeF4 + O2F2 ———->
Answer:

Question 29:
Draw the structures of the following:
(i) XeF2 (ii) BrF3
Answer:

Question 30:
Complete the following equations:
Ag + PCl5 ——–>
CaF2 + H2SO4 ——>
Answer:

Question 31:
Draw the structures of the following:
(i) XeF4 (ii) HCO4
Answer:

Question 32:
Complete the following equations:
(i) C+ conc. H2S04 ——->
(ii) XeF2 + H20 ——–>
Answer:

Question 33:
Draw the structures of the following:
(i) Xe03 (ii) H2SO4
Answer:

Question 34:
Name the two most important allotropes of sulphur. Which one of the two is stable at room temperature? What happens when the stable form is heated above 370 K?
Answer:
(i) Rhombic sulphur (α-sulphur)
(ii) Monoclinic sulphur (β-sulphur)
Rhombic sulphur is more stable at room temperature.
When Rhombic sulphur is heated above 370 K, it changes to monoclinic sulphur.

Question 35:
(i) Write the conditions to maximize the yield of H2S04 by contact process.
(ii) Why is Ka1< < Ka2 for H2S04 in water?
Answer:
(i) (a) High pressure, 2 bar (b) Temperature, 720
(c) V2O5, catalyst. (d) Excess of oxygen.
(ii) H2S04is a strong acid, therefore, its K is very high as it dissociates into H3O+ and HS04 almost completely.
The dissociation of HS04 to H30+ and SO-2 is slow, therefore, is much lower than Ka1

Short Answer Type Questions [II] [3 Marks]

Question 36:
36. (a) Draw the structures of the following molecules:
(i) XeOF4 (ii) H2S04
(b) Write the structural difference between white phosphorus and red phosphorus.
Answer:
(a) (i) Refer Ans. to Q. 13 (b) (ii).
(ii) Refer Ans. to Q.33 (ii).
(b) White phosphorus is monomeric, whereas red phosphorus is polymeric.

Question 37:
Account for the following:
(i) PCl5 is more covalent than PCl3.
(ii) Iron on reaction with HCl forms FeCl2 and not FeCl3.
(iii) The two O—O bond lengths in the ozone molecule are equal.
Answer:
(i) P5+ has more polarizing power than P3+, therefore, PCl5 is more covalent than PCl3 according to Fajan’s rule.
(ii) Refer Ans. to Q.20 (b) (i).
(iii) It is due to resonance.

Question 38:
(a) Draw the structures of the following:
(i) XeF2 (ii) BrF3
(b) Write the structural difference between white phosphorus and red phosphorus.
Answer:
(a) (i) Refer Ans. to Q. 15 (b) (ii).
(ii) Refer Ans. to Q.29 (ii).
(b) White phosphorus is monomeric, whereas red phosphorus is polymeric.

Question 39:
39. Account for the following:
(i) Bi(V) is a stronger oxidizing agent than Sb(V).
(ii) N—N single bond is weaker than P—P single bond. *’
(iii) Noble gases have very low boiling points.
Answer:
(i) Refer Ans. to Q.15 (a) (i).
(ii) It is due to more repulsion between valence electrons of smaller size of N atoms than P atoms.
(iii) It is due to weak van der Waals’ forces of attraction as these are non-polar.

Question 40:
40. (a) Draw the structures of the following compounds:
(i) XeF4 (ii) N205
(b) Write the structural difference between white phosphorus and red phosphorus.
Answer:

Question 41:
Account for the following:
(i) Sulphur in vapour form exhibits paramagnetic behaviour.
(ii) SnCl4 is more covalent than SnCl2.
(iii) H3PO2 is a stronger reducing agent than H3P03.
Answer:
(i) Sulphur exists as S2 molecule like O2 in vapour state and has two unpaired electrons. Therefore, it is paramagnetic.
(ii) Sn4+ has more polarising power than Sn2+ due to smaller size and higher charge. ,
(iii) It is because H3PO2 has two P—H bonds, whereas H3P03 has only one P—H bond

Question 42:
Give reasons for the following:
(i) (CH3)3P=0 exists but (CH3)3 N=0 does not.
(ii) Oxygen has less electron gain enthalpy with negative sign than sulphur.
(iii) H3P02 is a stronger reducing agent than H3P03.
Answer:
(i) It is because ‘N’ does not have d-orbitals, whereas ‘P’ has rf-orbitals.
(ii) It is due to more inter-electronic repulsion in smaller oxygep atoms than sulphur atoms.
(iii) Refer Ans. to Q.41 (iii).

Long Answer Type Questions [5 Marks]

Question 43:
(a) Account for the following:
(i) Bi is a strong oxidizing agent in the +5 state.
(ii) PCl5 is known but NCl5 is not known.
(iii) Iron dissolves in HCl to form FeCl2 and not FeCl3.
(b) Draw the structures of the following:
(i) XeOF4(ii) HClO4
Answer:
(a) (i) Bi5+can gain 2 electrons to form Bi3+which is more stable due to inert pair effect.
(ii) ‘P’ has ef-orbitals, whereas ‘N’ does not have cf-orbitals.
(iii) Refer Ans. to Q.20 (b) (i)
(b) (i) Refer Ans. to Q. 13 (b) (ii).
(ii) Refer Ans. to Q.31 (ii).

Question 44:
(a) Draw the structures of the following:
(i) H2S2O8(ii) Red P4
(b) Account for the following:
(i) Sulphur in vapour state exhibits paramagnetism.
(ii) Unlike xenon, no distinct chemical compound of helium is known.
(iii) H3P02 is a stronger reducing agent than H3P03.
Answer:

(b) (i) Refer Ans. to Q.41 (i).
(ii) Helium (He) has the highest ionisation enthalpy and least polarising power due to smaller size as compared to Xe, therefore, it does not form chemical compound.
(iii) Refer Ans. to Q.41 (iii).

Question 45:
What is the covalency of nitrogen in N2O5?
Answer:

2013

Very Short Answer Type Questions [1 Mark]

Question 46:
Very Short Answer Type Questions [1 Mark]

What inspired N. Bartlett for carrying out reaction between Xe and PtF6
Answer:
The ionisation enthalpy of O2 and Xe are nearly same, therefore, he prepared Xe+[PtF6]- like 02+ [PtF6]-

Question 47:
What is the basicity of H3P03 and why?
Answer:
It is dibasic acid because it has two replaceable hydrogen attached with oxygen.

Question 48:
Name two poisonous gases which can be prepared from chlorine gas.
Answer:
(i) Chloropicrin (ii) Phosgene (iii) Mustard gas

Question 49:
Which aerosol depletes ozone layer?
Answer:
Chlorofluorocarbons (CFCs).

Short Answer Type Questions [I] [2 Marks]

Question 50:
What happens when
(i)PCl5 is heated? (ii) H3P03 is heated?
Write the reactions involved.
Answer:

Question 57:
Give reasons for the following:
(i) Though nitrogen exhibits +5 oxidation state, it does not form pentahalide.
(ii) Electron gain enthalpy with negative sign of fluorine is less than that of chlorine.
(iii) The two oxygen-oxygen bond lengths in ozone molecule are identical.
Answer:
(i) It is because it does not have cf-orbitals.
(ii) It is due to more inter electronic repulsion in smaller size of‘F’ as compared to Cl.
(iii) It is due to resonance.

Question 58:
Give reasons for the following:
(i) Oxygen is a gas but sulphur is a solid.
(ii) O3 acts as a powerful oxidising agent.
(iii) BiH3 is the strongest reducing agent amongst all the hydrides of Group 15 elements.
Answer:
(i) It is because oxygen is diatomic and has less intermolecular forces of attraction, whereas sulphur is octa-atomic (S8) and has more intermolecular forces of attraction.
(ii) It is because Os has low bond dissociation energy and it is more reactive. Therefore, it liberates nascent oxygen easily.
O3 ———> O2+ [O]
(iii) Refer Ans. to Q.20 (b) (iii).

Question 59:
Draw the structures of the following molecules:
(i) N2O5 (ii) H3PO2 (iii) XeF6
Answer:

Question 60:
Account for the following:
(i) White phosphorus is more reactive than red phosphorus.
(ii) SnCl4 is more covalent than SnCl2.
(iii) O3 is a powerful oxidising agent.
Answer:
(i) It is because white phosphorus is monomeric and has less bond dissociation energy, whereas red phosphorus is polymeric and has more bond dissociation energy.
(ii) Refer Ans. to Q.41 (ii).
(iii) Refer Ans. to Q.58 (ii).

Question 62:
62. Draw the structures of the following molecules:
(i) PCl3(ii) H4P207 (iii) CIF3
Answer:

Question 63:
Draw the structures of the following:
(i) Solid PCl5 (ii) H2S2O8 (iii) Xe0
Answer:

Long Answer Type Questions [5 Marks]

Question 64:
(a) Give reasons for the following:
(i) Bond enthalpy of F2 is lower than that of Cl2.
(ii) PH3 has lower boiling point than NH?.
(b) Draw the structures of the following molecules:
(i) BrF3 (ii) (HP03)3 (iii) XeF4
Answer:
(a) (i) It is due to more repulsion between valence electrons of F than Cl due to exceptionally small size.
(ii) PH3 molecules are not associated with H-bonding, whereas NH3 molecules are associated with H-bonding.

Question 65:
(a) Account for the following:
(i) Helium is used in diving apparatus.
(ii) Fluorine does not exhibit positive oxidation state.
(iii) Oxygen shows catenation behaviour less than sulphur.
(b) Draw the structures of the following molecules:
(i) XeF2 (ii) H2S2O8
Answer:
(a) (i) It is less soluble in blood than nitrogen and does not cause bends (pain) in body under the pressure of water.
(ii) Fluorine is the most electronegative element and does not have ri-orbitals.
(iii) It is due to more repulsion between valence electrons of two oxygen atoms in 0-0 than S-S due to smaller atomic size. Therefore, 0-0 bonds is weaker than S-S bonds.
(b) (i) Refer Ans. to Q.15 (b) (ii).
(ii) Refer Ans. to Q.44 (a) (i).

2012

Very Short Answer Type Questions [1 Mark]

Question 66:
Which one of PCl-4 and PCl4 is not likely to exist and why?
Answer:
PCl4 does not exist because octet of ‘P’ is not complete and it is unstable.

Question 67:
Of PH3 and H2S which is more acidic and why?
Answer:
H2S is more acidic due to lower bond dissociation enthalpy. ‘S’ is more electronegative than phosphorus.

Question 68:
Although the H-bonding in hydrogen fluoride is much stronger than that in water yet water has a much higher boiling point than hydrogen fluoride. Why ?
Answer:
It is because the extent of hydrogen bonding is more in H2O (can form four H-bonds) than HF (can form two H-bonds), therefore, H20 has higher boiling point than HF.

Question 69:
Draw the structure of XeF6.
Answer:
Refer Ans. to Q.51 (i).

Question 70:
Despite lower value of its electron gain enthalpy with negative sign, fluorine (F2) is a stronger oxidising agent than Cl2.
Answer:
It is due to higher standard reduction potential of F2 which is due to low bond dissociation energy of F—F bond because of repulsion among small size F atoms, high electron gain enthalpy and highest hydration enthalpy.

Question 71:
Which is a stronger reducing agent, SbH3 or BiH3, and why?
Answer:
BiH3 is a stronger reducing agent because it has low bond dissociation energy than SbH3 due to longer bond length.

Question 72:
What is the basicity of H3P02 acid and why?
Answer:
It is monobasic acid because it has only one replaceable hydrogen atom

Question 73:
Complete the following chemical equation:
NH4Cl(aq) + NaNO2(aq) ———>
Answer:
NH4Cl(aq) + NaNO2(aq) ——> NaCl(aq) + N2(g) + 2H2O(l)

Question 74:
Which is a stronger acid in aqueous solution, HF or HCl, and why?
Answer:
HCl because bond dissociation energy of H—Cl is lower than HF.

Question 75:
Which is more acidic and why, H20 or H2S?
Answer:
H2S is more acidic than H20 due to low bond dissociation enthalpy due to longer bond length than H20.

Short Answer Type Questions [I] [2 Marks]

Question 76:
Explain the following giving an appropriate reason in each case.
(i) O2 and F2 both stabilize higher oxidation states of metals but 02 exceeds F2 in doing so.
(ii) Structures of Xenon fluorides cannot be explained by Valence Bond approach
Answer:
(i) It is due to higher lattice energy of oxides as compared to fluorides as oxide ion is dinegative, whereas fluoride ion is mononegative.
(ii) It is because Xenon (Xe) is a noble gas and has lone pair of electrons in its d-orbitals. The size of bd orbital in Xe is large enough for effective overlapping, bp and bd orbitals of Xe differ by 960 kj mol-1, that is, sp5/ hybridisation, contribution of bd orbital is objectionable. SO, valence bond approach is not suitable.

Question 77:
Explain the following facts giving appropriate reason in each case:
(i) NF3 is an exothermic compound whereas NCl3 is not.
(ii) All th bonds in SF4 are not equivalent.
Answer:
(i) It is because F2 is stronger oxidising agent than Cl2, therefore, NF3 is exothermic compound, whereas NCl3 is endothermic as N—F bond is stronger than N—Cl bond.
(ii) In SF4, bonds are in different planes, therefore, they are not equivalent.

Question 78:
Explain the following:
(i) The chemical reactivity of nitrogen is much less than that of phosphorus.
(ii) SF6 is kinetically inert.
Answer:
(i) It is because nitrogen has triple bond which has high bond dissociation enthalpy than single bond in phosphorus.
(ii) SF6 is stearically protected, therefore, kinetically inert.

Question 79:
Draw the molecular structures of the following species:
(i) H3PO3 (ii) BrF3
Answer:
(i) Refer Ans. to Q.53 (ii).
(ii) Refer Ans. to Q.29 (ii).

Question 80:
Draw the molecular structures of the following species:
(i) H2S2O8 (ii) XeF2
Answer:
(i) Refer Ans. to Q. 13 (b) (i).
(ii) Refer Ans. to Q.15 (b) (ii).

Question 81:
State a reason for each of the following statements:
(i) Fluorine never exhibits any positive oxidation state.
(ii) Helium does not form any real chemical compounds.
Answer:
(i) Fluorine is the most electronegative and does not have c?-orbitals, therefore, it does not show +ve oxidation state.
(ii) Helium has highest ionization enthalpy, so, it does not form any real compound.

Short Answer Type Questions [II] [3 Marks]

Question 82:

Answer:

Question 83:

Answer:

Question 84:

Answer:

Long Answer Type Questions [5 Marks]

Question 85:
(
a) Draw the molecular structures of the following compounds:
(i) XeF6 (ii) H2S2O8
(b) Explain the following observations:
(i) The molecules NH3 and NF3 have dipole moments which are of opposite direction.
(ii) All the bonds in PCl5 molecules are not equivalent.
(iii) Sulphur in vapour state exhibits paramagnetism.
Answer:
(a) (i) Refer Ans. to Q.51 (i).
(ii) Refer Ans. to Q. 13 (b) (ii).
(b) (i) “N’ is more electronegative than ‘H’ but F is more electronegative than N.

(ii) It is because these are not in the same plane. Axial bonds are longer and weaker than Equatorial bonds due to more repulsion.
(iii) Refer Ans. to Q.41 (i).

Question 86:
(a) Complete the following reaction
(i) XeF4 + SbF5 ——–> (ii) Cl2 + F2 (excess) ——>
(b) Explain each of the following:
(i) Nitrogen is much less reactive than phosphorus.
(ii) The stability of +5 oxidation state decreases down in group 15.
(iii) The bond angles (O-N-O) are not of the samevalue inN02- and N02+
Answer:
(a) (i) XeF 4 + SbF- ——> [XeF3]+ |SbF6]-
(ii) Cl2 + 3F2(excess) ——-> 2ClF3(g)
(b) (i) Is is due to high bond dissociation enthalpy of N = N (triple bond) than single bond in P4.
(ii) It is due to inert pair effect.
(iii) It is due to more repulsion in N02 than in N02 due to presence of lone pair of electron.

Question 87:
(a) Complete the following chemical equations:
(i) Cu + HNO3 (dilute) ——->
(ii) XeF4 + O2F2 ———->
(b) Explain the following observations:
(i) Phosphorus has greater tendency for catenation than nitrogen.
(ii) Oxygen is a gas but sulphur is a solid.
(iii) The halogens are coloured. Why ?
Answer:


(b) (i) It is because N—N bond is weaker than P—P bond due to more interelectronic repulsion due to smaller size.
(ii) Refer Ans. to Q.58 (i).
(iii) They are coloured because their molecules absorb light from visible region and outer electrons get excited to higher energy level. When they come back to lower energy level, they radiates complementary colours.

Question 88:

Answer:

Question 89:

Answer:

(ii) Refer Ans. to Q.83 (in).
(b) (i) Refer Ans. to Q.59 (ii).
(ii) Refer Ans. to Q.51 (ii).
(iii) Refer Ans. to Q. 13 (b) (ii).

Question 90:
(a) Draw the molecular structures of the following compounds:
(i) N2O5 (ii) XeOF4
(b) Explain the following observations:
(i) Sulphur has a greater tendency for catenation than oxygen.
(ii) ICl is more reactive than I2.
(iii) Despite lower value of its electron gain enthalpy with negative sign, fluorine (F2) is a stronger oxidising agent than Cl2.
Answer:
(a) (i) Refer Ans. to Q.40 (a) (ii).
(ii) Refer Ans. to Q.13 (b) (ii).
(b) (i) Refer Ans. to Q.65 (iii).
(ii) It is due to low bond dissociation enthalpy due to less effective overlapping in ICl than I2.
(iii) Refer Ans. to Q.70.

Question 91:
(a) Draw the structures of the following compounds:
(i) H3P02 (ii) CIF3
(b) Explain the following observations:
(i) Nitrogen is much less reactive than phosphorus.
(ii) Despite having greater polarity, hydrogen fluoride boils at a lower temperature than water.
(iii) Sulphur has greater tendency for catenation than oxygen in the same group.
Answer:
(a) (i) Refer Ans. to Q.59 (ii).
(ii) Refer Ans. to Q. 17 (b) (i).
(b) (i) Refer Ans. to Q.86 (b) (i).
(ii) It is because extent of H-bonding is more in H20 than HF because each water molecule can form four H-bonds.
(iii) Refer Ans. to Q.65 (in).

Question 92:
(a) Draw the structures of the following molecules:
(i) N2O5 (ii) HClO4
(b) Explain the following observations:
(i) H2S is more acidic than H2O.
(ii) Fluorine does not exhibit any positive oxidation state.
(iii) Helium forms no real chemical compound
Answer:
(a) (i) Refer Ans. to Q.40 (a) (ii).
(ii) Refer Ans. to Q.31 (ii).
(b) (i) It is because bond dissociation energy of H—S bond is less than H—O bond due to longer bond length.
(ii) Refer Ans. to Q.65 (a) (ii).
(iii) Refer Ans. to Q.81 (ii).

2011

Very Short Answer Type Questions [1 Marks]

Question 93:
Arrange F2, Cl2, Br2 and I2 in the order of increasing bond dissociation enthalpy.
Answer:
I2 < F2 < Br2 < Cl2

Question 94:
Draw the structure of XeF2 molecule.
Answer:
Refer Ans. to Q. 15 (b) (ii).

Question 95:
Draw the structure of XeF4molecule.
Answer:
Refer Ans. to Q.17 (b) (ii).

Question 96:
Draw the structure of BrF3 molecule.
Answer:
Refer Ans. to Q.29 (ii).

Short Answer Type Questions [I] [2 Marks]

Question 97:
State reasons for each of the following:
(i) The N—O bond in NO2– is shorter than the N—O bond in N03.
(ii) SF6 is kinetically an inert substance.
Answer:

Question 98:
State reasons for each of the following:
(i) All the P—Cl bonds in PCl5 molecule are not equivalent.
(ii) Sulphur has greater tendency for catenation than oxygen.
Answer:
(i) Refer Ans. to Q.85 (b) (ii).
(ii) Refer Ans. to Q.65 (iii).

Question 99:
How would you account for the following:
(i) The following order of increase in strength of acids:
PH3 < H2S < HCl
(ii) The oxidising power of oxoacids of chlorine follows the order:
HCl04 < HClO3 < HClO2 < HCIO
Answer:
(i) It is because greater the difference in electronegativity, more will be the polarity and hence, more will be acidic character.
(ii) It is because HCIO is least stable and gives [O] most easily, whereas tendency to give oxygen in HCl04 is least where number of oxygen are maximum and oxidising power is least.

Question 100:
Draw the structures of the following molecules:
(i) XeOF4 (ii) H0Cl02
Answer:

Question 101:
Complete the following reaction equation:
(i) XeF4 + H20 —–> (ii) I2 + H20 + Cl2 ——>
Answer:

Question 102:
Complete the following reaction equation:
(i) XeF6 + H20 —–>
(ii) FeS04 + H2S04 + Cl2 ——->
Answer:

Question 103:
Explain giving a reason each for the following situations:
(i) In aquesous medium HCl is a stronger acid than HF.
(ii) White phosphorus is more reactive than red phosphorus.
Answer:
(i) Refer Ans. to Q.74.
(ii) Refer Ans. to Q.60 (i).

Question 104:
Complete the following reaction equation:
(i) XeF2(s) + H2O(l) ———->
(ii) NaOH(cold & dilute) + Cl2 ——>
Answer:
(i) Refer Ans. to Q.84 (iii).
(ii) Refer Ans. to Q.83 (ii).

Short Answer Type Questions [II] [3 Marks]

Question 105:
How would you account for the following:
(i) H2S is more acidic than H2O.
(ii) The N—O bond in N02 is shorter than the N—O bond in N03.
(iii) Both O2 and F2 stabilize high oxidation states but the ability of oxygen to stabilize the higher oxidation state exceeds that of fluorine
Answer:
(i) It is because bond dissociation energy of H—S bond is less than H—O bond due to longer bond length.
(ii) Refer Ans. to Q.97 (i).
(iii) It is because oxygen can gain two electrons, therefore, lattice energy can overcome ionisation energy of the metal to show higher oxidation state.

Question 106:
How would you account for the following:
(i) NF3 is an exothermic compound but NCl3 is not.
(ii) The acidic strength of compounds increases in the order:
PH3 < H2S < HCl
(iii) SF6 is kinetically inert.
Answer:
(i) Refer Ans. to Q.77 (i).
(ii) It is because bond dissociation energy of HCl is lower than H2S which is lower than PH3 due to greater polarity in HCl than H2S and H2S has more polarity than PH3, due to more difference in electronegativity.
(iii) It is because SF6 is stearically protected, therefore, an inert substance.

Question 107:
(a) Mention the optimum conditions for the industrial manufacture of ammonia by Haber’s process.
(b) Explain the following giving appropriate reasons:
(i) Sulphur vapour exhibits paramagnetic behaviour:
(ii) Red phosphorus is less reactive than white phosphorus
Answer:
(i) High pressure, 200 atm.
(ii) Temperature —700 K.
(iii) Catalyst, such as Iron oxide with small amount of K2O and Al2O3.
(b) (i) Refer Ans. to Q.41 (i).
(ii) Refer Ans. to Q.60 (i).

Question 108:
Draw the structures of the following molecules:
(i) NF3 (ii) H2S2O8 (iii) H3PO3
Answer:

Question 109:
Complete the following chemical equations:
(i) P4 + S02Cl2 ——>
(it) Fe3+ + S02 + H20 ——–>
(iii) XeF6 + H20(excess) ———>
Answer:
(i) Refer Ans. to Q.88 (a) (i).
(ii) Refer Ans. to Q.84 (ii).
(ii) Refer Ans. to Q.88 (a) (ii).

Question 110:
Account for the following:
(i) Nitrogen does not form pentahalides.
(ii) The two oxygen-oxygen bond lengths in ozone (O3) molecule are same.
(iii) ICl is more reactive than I2.
Answer:
(i) It is because nitrogen does not have vacant d-orbitals.
(ii) Refer Ans. to Q.57 (iii).
(iii) Refer Ans. to Q.90 (b) (ii).

Question 111:
Account for the following:
(i) Ammonia is more basic than phosphine.
(ii) Elements of Group 16 generally show lower value of first ionisation enthalpy compared to the elements in the corresponding periods of Group 15.
(iii) Electron gain enthalpy with negative sign for fluorine is less than that for chlorine.
Answer:
(i) It is because ‘N’ is smaller in size and lone pair of electron is readily available for protonation.
(ii) It is because group 15 elements have half filled p-orbitals which are more stable, therefore, they have higher ionisation enthalpy than group 16 elements.
(iii) Refer Ans. to Q.57 (ii).

Question 112:
Account for the following:
(i) PCl5 can act as an oxidising agent but not as a reducing agent.
(ii) Dioxygen is a gas but sulphur is a solid.
(iii) Halogens are coloured.
Answer:
(i) It is because ‘P’ has +5 oxidation in PCl5. It cannot show higher oxidation state, therefore, it can not act as reducing agent. It can act as oxidising agent as it can gain electron to show lower oxidation state.
(ii) Refer Ans. to Q.58 (i).
(iii) It is because they absorb light from visible region and radiate complementary colour.

Question 113:
Account for the following:
(i) NH3acts as a good ligand.
(ii) H2S is more acidic than water.
(iii) Fluorine forms the largest number of interhalogen compounds amongst the halogens.
Answer:
(i) NH3 acts as a good ligand due to presence of lone pair of electron which it can readily donate.
(ii) Refer Ans. to Q.75.
(iii) It is because fluorine is the most electronegative and strongest oxidising agent.

Question 114:
Account for the following:
(i) BiCl3 is less covalent than PCl3.
(ii) O3 acts as a powerful oxidising agent.
(iii) F2 is a stronger oxidising agent than Cl2.
Answer:
(i) It is because ionisation enthalpy of Bi is lower than phosphorus, therefore, Bi forms ionic BiCl3 whereas PCl3 is covalent.
(ii) Refer Ans. to Q.58 (ii).
(iii) It is because F2 has highest standard reduction potential, higher than Cl2.

Question 115:
Account for the following:
(i) BiH3 is the strongest reducing agent amongst all the hydrides of Group 15 elements.
(ii) Ka2<< Ka1 for H2S04 in water.
(iii) Fluorine forms only one oxoacid, HOF.
Answer:
(i) It is due to low bond dissociation energy which is due to longer bond length.
(ii) Refer Ans. to Q.35 (ii).
(iii) Fluorine is the most electronegative and small in size, therefore, it does not show positive oxidation state. It forms only HOF at -40°C.

Question 116:
How would you account for the following:
(i) NCl3 is an endothermic compound while NF3 is an exothermic one.
(ii) XeF2 is a linear molecule without a bend.
(iii) The electron gain enthalpy with negative sign for fluorine is less than that for chlorine, still fluorine is a stronger oxidising agent than chlorine.
Answer:
(i) Refer Ans. to Q.77 (i).
(ii)XeF2 is linear because it has two bonded pair which are at 180° where as 3 lone pair of electrons are at 120°.
(iii) Refer Ans. to Q.70.

Question 117:
How would you account for the following:
(i) The electron gain enthalpy with negative sign is less for oxygen than that for sulphur.
(ii) Phosphorus shows greater tendency for catenation than nitrogen.
(iii) Fluorine never acts as the central atom in polyatomic inter-halogen
compounds.
Answer:
(i) It is due to more interelectronic repulsion in oxygen than sulphur due to small size of oxygen atom.
(ii) P—P bond is stronger than N—N bond due to less repulsion between valence electrons.
(iii) ‘F’ does not show higher oxidation state due to absence of ef-orbitals.

Long Answer Type Questions [5 Marks]

Question 118:
(a) Explain the following:
(i) NF3 is an exothermic compound whereas NCl3 is not.
(ii) F2 is most reactive of all the four common halogens.
(b) Complete the following chemical equations:
(i) C + H2S04 (conc.) ———>
(ii) P4 + NaOH + H2O ———>
(iii) Cl2 + F2 ——–>
Answer:
(a) (i) It is because NF3 is more stable due to stronger N—F bond than NCl3 because F2 is stronger oxidising agent than Cl2.
(ii) It is due to low bond dissociation energy, high hydration energy and high electron affinity.

Question 119:
(a) Account for the following:
(i) The acidic strength decreases in the order HCl > H2S > PH3
(ii) Tendency to form pentahaUdes decreases down the group in group 15 of the periodic table.
(b) Complete the following chemical equations:
(i) P4 + SO2Cl2 ———>
(ii) XeF2 + H2O ——->
(iii) I2 + HNO3 (conc.) ———–>
Answer:
(a) (i) It is because bond dissociation energy of HCl is lower than H2S which
is lower than PH3. This is due to greater polarity in HCl than H2S and H2S has more polarity than PH3, due to more difference in electronegativity.
(ii) It is because inert pair effect, e.g. Bi3+ is more stable than Bi5+.

Question 120:

Answer:

Question 121:
(a) What happens when
(i) chlorine gas is passed through a hot concentrated solution of NaOH?
(ii) sulphur dioxide gas is passed through an aqueous solution of a Fe (III) salt?
(b) Answer the following:
(i) What is the basicity of H3P03 and why?
(ii) Why does fluorine not play the role of a central atom in interhalogen compounds?
(iii) Why do nobel gases have very low boiling points?
Answer:


(b) (i) Refer Ans. to Q.47.
(ii) Refer Ans. to Q. 117 (iii).
(iii) It is due to weak van der Waals’ force of attraction between atoms of noble gases.

Question 122:

(b) How would you account for the following?
(i) The value of electron gain enthalpy with negative sign for sulphur is higher than that for oxygen.
(ii) NF3 is an exothermic compound but NCl3 in endothermic compound.
(iii) ClF3 molecular has a T-shaped structure and not a trigonal planar one.

Answer:
(a) (i) Refer Ans. to Q.89 (a) (i).
(ii) Refer Ans. to Q.88 (a) (ii).
(b) (i) Refer Ans. to Q.117 (i).
(ii) Refer Ans. to Q.77 (i).
(iii) ClF3 has two lone pair and 3 bonded pair, therefore, it is T-shaped and not a trigonal planar.

Question 123:
(a) Complete the following chemical equations:
(i) P4 + S02Cl2 ——-> (ii) XeF4 + H20 ——–>
(b) Explain the following observations giving appropriate reasons:
(i) The stability of + 5 oxidation state decreases down the group in group 15 of the periodic table.
(ii) Solid phosphorus pentachloride behaves as an ionic compound.
(iii) Halogens are strong oxidising agents.
Answer:
(a) (i) Refer Ans. to Q.119 (b) (i).
(ii) Refer Ans. to Q. 120 (b) (iii).
(b) (i) Refer Ans. to Q.86 (b) (ii).
(ii) It exists as [PCl4]+ [PCl6]- in solid state, therefore, it behaves like ionic compound.
(iii) It is because they have high electron gain enthalpies, therefore, they can gain electron easily.

2010

Very Short Answer Type Questions [1 Mark]

Question 124:
Why does N02 dimerise?
Answer:
Refer Ans. to Q.22.

Question 125:
What is the oxidation number of phosphorus in H3PO2 molecule?
Answer:


Question 126:
Draw the structure of 03 molecule.
Answer:

Question 127:
Fluorine does not exhibit any positive oxidation state. Why?
Answer:
It is because it is most electronegative element and best oxidising agent.

Question 128:
Nitrogen is relatively inert as compared to phosphorus. Why?
Answer:
It is due to presence of triple bond in nitrogen (N==N), which has high bond dissociation energy as compared to single (P-P) bond.

Question 129:
Which is a stronger acid in aqueous solution, HCl or HI, and why?
Answer:
HI is stronger acid than HCl in aqueous solution because it has lower bond dissociation energy.

Question 130:
What is the covalency of nitrogen in N2O5?
Answer:
It is four.

Question 131:
Why are pentahalides of a metal more covalent than its trihalides?
Answer:
It is because pentavalent metal ion has higher polarising power than trivalent metal ion

Question 132:
Why is BiH3 the strongest reducing agent amongst all the hydrides of Group 15 elements?
Answer:
It is due to its lowest bond dissociation energy due to longer bond length.

Short Answer Type Questions [I] [2 Marks]

Question 133:
Draw the structures of white phosphorus and red phosphorus. Which one of these two types of phosphorus is more reactive and why?
Answer:

Question 134:
Draw the structural formulae of molecules of following compound:
(i) BrF3 and (ii) XeF4
Answer:
(i) Refer Ans. to Q.29 (ii).
(ii) Refer Ans. to Q.17 (b) (ii).

Question 135:
Complete the following chemical reaction equations:

Answer:
(i) Refer Ans. to Q.119 (b) (iii).
(ii) Refer Ans. to Q.120 (b) (i)

Short Answer Type Questions [II] [3 Marks]

Question 136:
Draw the structural formulae of the following compounds:
(i) H4P2O5 (ii) XeF4
Answer:

Question 137:

Answer:
(i) Refer Ans. to Q.83 (ii).
(ii) Refer Ans. to Q.102 (i).

Question 138:
Draw the structures of the following molecules:
(i) BrF3 (it) XeOF4
Answer:
(i) Refer Ans. to Q.29 (ii).
(ii) Refer Ans. to Q.13 (b) (ii).

Question 139:
Draw the structure of O2 and S8 molecules Ozone (O3):
Answer:

Question 140:
Draw the structures of the following molecules:
(i) XeF2 (ii) HClO4
Answer:
(i) Refer Ans. to Q.15 (b) (ii).
(ii) Refer Ans. to Q.31 (ii).

Question 141:
Draw the structure and predict the shape of (i) XeO3 and (ii) BrF3
Answer:
(i) Refer Ans. to Q.33 (i).
(ii) Refer Ans. to Q.29 (ii).

Short Answer Type Questions [II] [3 Marks>

Question 142:
Give reasons for the following:
(i) N2 is not particularly reactive.
(ii) Halogens are strong oxidising agents.
(iii) Sulphur hexafluoride is less reactive than sulphur tetrafluroide.
Answer:
(i) It is due to high bond dissocadon energy which is due to presence of triple bond.
(ii) Halogens can gain electron easily, have high electron affinity and reduction potential, therefore, they are strong oxidising agents.
(iii) SF6 is stearically protected, therefore, less reactive than SF4 which is not stearically protected.

Question 143:
Explain the following observations giving appropriate reasons:
(i) The stability of +5 oxidation state decreases down the group in group 15 of the periodic table.
(ii) Solid phosphorus pentachloride behaves as an ionic compound.
(iii) Halogens are strong oxidizing agents.
Answer:
(i) It is due to inert pair effect, +3 oxidation state becomes more stable than + 5.
(ii) It exists as [PCl4]+ [PCl6]- in solid state, therefore, it behaves like ionic compound.
(iii) Halogens can gain electron easily and have high standard reduction potential, therefore, good oxidising agents.

Long Answer Type Questions [5 Marks]

Question 144:
(a) Complete the following chemical reaction equations:
(i) HgCl2 (aq) + PH3 (g) ——->
(ii) Si02 (g) + HF (g) ———–>
(b) Explain the following observations:
(i) Sulphur in vapour state exhibits paramagnetic behaviour.
(ii) The stability of +3 state increases down the group in group 15 of the periodic table.
(iii) XeF2 has a linear shape and not a bent structure.
Answer:
(a) (i) 3HgCl2 (aq) + 2PH3 (g) ——-> Hg3P2 (5) + 6HCl(aq)
(ii) Si02 (5) + 6HF (g) ———> H2SiF6 (s) + 2H20(l)
(b) (i) It is due to presence of two unpaired electrons in S2 like in O2 in vapour state.
(ii) It is due to inert pair effect.
(iii) It is due to presence of two bonded pairs and three lone pairs of electrons.

Question 145:
(a) Complete the following chemical reaction equations:
(i) AgCl (s) + NH3 (aq) ——->
(ii) P4(s) + NaOH(aq) + H20(l) ——–>
(b) Explain the following observations:
(i) H2S is less acidic than H2Te.
(ii) Fluorine is a stronger oxidising agent than chlorine.
(iii) Noble gases are the least reactive elements
Answer:
(a) (i) AgCl(s) + 2NH3(aq) ——–> [Ag(NH3)2]+Cl-(aq)
(ii) P4(s) + 3NaOH(aq) + 3H2O(l) ——–> 3NaH2P02(s) + PH3(g)
(b) (i) It is because bond dissociation energy of H-Te bond is less than H-S bond due to longer bond length.
(ii) It is due to higher standard reduction potential, low bond dissociation energy, high electron affinity and higher enthalpy of hydration.
(iii) It is due to stable electronic configuration, i.e. their octet is complete except in He which has duplet, i.e. 1st shell is complete having 2 electrons.

2009

Very Short Answer Type Questions [1 Mark]

Question 146:
Why is Bi (V) a stronger oxidant than Sb(V)?
Answer:
Refer Ans. to Q. 15 (a) (i).

Question 147:
Why is red phosphorus less reactive than white phosphorus?
Answer:
Refer Ans. to Q.60 (i).

Question 148:
Assign a reason for each of the following statements:
Phosphorus (P4) is more reactive than nitrogen (N2).
Answer:
It is due to single bond in phosphorus which has less bond dissociation energy as compared to nitrogen which has triple bond (N=N) has high bond dissociation energy, so, nitrogen is unreactive

Question 149:
Which one has higher electron gain enthalpy with negative sign, sulphur or oxygen?
Answer:
Sulphur.

Short Answer Type Questions [I] [2 Marks]

Question 150:

Answer:

Question 151:
State reasons for each of the following:
(i) All the P—Cl bonds in PCl5 molecule are not equivalent.
(ii) Sulphur has greater tendency for catenation than oxygen.
Answer:
(i) Refer Ans. to Q.85 (b) (ii).
(ii) Refer Ans. to Q.65 (iii).

Question 152:
Answer the following:
(i) Which neutral molecule would be isoelectronic with CO-?
(ii) Of Bi(V) and Sb(V) which may be a stronger oxidising agent and why?
Answer:
(i) OF2 and ClF are neutral molecules isoelectronic with CIO-.
(ii) Bi(V) is stronger oxidising agent due to inert pair effect as Bi(III) is more stable as compared to Sb(III).

Question 153:
Complete the following chemical reaction equations:
(i) XeF2 + H20 ——> (ii) PH3 + HgCl2 ——>
Answer:
(i) Refer Ans. to Q.84 (ii).
(ii) Refer Ans. to Q.83 (i).

Question 154:
Draw the structural formulae of molecules of following compound:
(i) BrF3 and (ii) XeF4
Answer:
(i) Refer Ans. to Q.29 (ii).
(ii) Refer Ans. to Q. 17 (b) (ii).

Short Answer Type Questions [III] [3 Marks]

Question 155:
Account for the following:
(i) NH3 is a stronger base than PH3.
(ii) Sulphur has a greater tendency for catenation than oxygen.
(iii) Bond dissociation energy of F2 is less than that of Cl?
Answer:
(i) Refer Ans. to Q.111 (i).
(ii) Refer Ans. to Q.65 (ii).
(iii) It is due to interelectronic repulsion between valence electrons of smaller size ‘F’ atom than Cl.

Question 156:
Explain the following situations:
(i) In the structure of HNO3 molecule, the N—O bond (121 pm) is shorter than N—OH bond (140 pm).
(ii) SF4 is easily hydrolysed whereas SF6 is not easily hydrolysed.
(iii) XeF2 has a straight linear structure and not a bent angular structure
Answer:
(i) There is double bond character in N—O bond, therefore, it is shorter than N—OH bond which has purely single bond.
(ii) SF4 is easily hydrolysed because it is unstable due to repulsion between ‘F’ atoms and also due to vacant d-orbitals, whereas SF6 is exceptionally stable due to steric reason.
(iii) XeF2 has sp3d hybridisation with three lone pairs of electrons at corners of equilateral triangle, it has linear shape to have minimum repulsion and maximum stability.

Question 157:
Explain the following observations:
(i) Fluorine does not exhibit any positive oxidation state.
(ii) The majority of known noble gas compounds are those of Xenon.
(iii) Phosphorus is much more reactive than nitrogen.
Answer:
(i) Refer Ans. to Q.127.
(ii) Xe has lowest ionisation enthalpy and high polarizing power* Therefore, it can form compounds easily.
(iii) Refer Ans. to Q.148.

Question 158:
Draw the structures of the following molecules:
(i) BrF3
(ii) H2S2O 7
Answer:
(i) Refer Ans. to Q.29 (ii).
(ii) Refer Ans. to Q.20 (a) (ii).

Question 159:
Draw the structures of the following molecules:
(i) XeF4 (ii) H2S2O7
Answer:
(i) Refer Ans. to Q. 17 (b) (ii).
(ii) Refer Ans. to Q.20 (a) (ii).

Long Answer Type Questions [5 Marks]

Question 160:
(a) Complete the following chemical reaction equations:
(i) HgCl2 (aq) + PH3 (g) ——->
(ii) SiO2 (g) + HF (g) ——->
(b) Explain the following observations:
(i) Sulphur in vapour state exhibits paramagnetic behaviour.
(ii) The stability of +3 state increases down the group in group 15 of the
periodic table.
(iii) XeF2 has a linear shape and not a bent structure.
Answer:
(a) (i) 3HgCl2 (aq) + 2PH3 (g) ——-> Hg3P2 (s) + 6HCl (aq)
(ii) SiO2 (s) + 6HF (g) ——-> H2SiF6 (s) + 2H20(l)
(b) (i) It is due to presence of two unpaired electrons in S2 like in O2 in vapour state.
(ii) It is due to inert pair effect.
(iii) It is due to presence of two bonded pairs and three lone pairs of electrons.

Question 161:
(a) Draw the structures of the following:
(i) H2S2O8 (ii) HClO4
(b) How would you account for the following:
(i) NH3 is a stronger base than PH3.
(ii) Sulphur has a greater tendency for catenation than oxygen.
Answer:
(a) (i) Refer Ans. to Q. 13 (b) (i).
(ii) Refer Ans. to Q.31 (ii).
(b) (i) Refer Ans. to Q.111 (i).
(ii) Refer Ans. to Q.65 (iii).

Question 162:
(a) Draw the structures of the following:
(i) XeF4(ii) H2S2O7
(b) Explain the following observations:
(i) Phosphorus has a greater tendency for catenation than nitrogen.
(ii) The negative value of electron gain enthalpy is less for fluorine than that for chlorine.
(iii) Hydrogen fluoride has a much higher boiling point than hydrogen chloride.
Answer:
(a) (i) Refer Ans. to Q. 17 (b) (ii).
(ii) Refer Ans. to Q.20 (a) (ii).
(b) (i) It is because P—P single bond is stronger than the single N—N bond.
(ii) It is because there is more interelectronic repulsion between valence electrons in ‘F’ atoms as compared to ‘CP atoms.
(iii) It is because HF molecules are associated with intermolecular H-bonding while HCl is not, that is why, HF is liquid and has higher boiling point than HCl which is a gas.

Question 163:
(a) Complete the following reaction equations:
(i) PCl5 + H20 (excess) —–>
(ii) F2 + H20 ———>
(b) Explain the following observations:
(i) No distinct chemical compound of helium is known.
(ii) Phosphorus has a greater tendency for catenation than nitrogen.
(iii) In solution of H2S04 in water, the second dissociation constant K02, is less than the first dissociation constant K .
Answer:


(b) (i) It is because helium is smallest in size, has higher ionisation energy and therefore, due to stable electronic configuration. It has least polarizing power.
(ii) It is because of strong bond strength of P-P than N-N bond.
(iii) Refer Ans. to Q.35 (ii).

Question 164:
(a) Complete the following reaction equations:
(i) P4 + NaOH + H20 ———>
(ii) Cu + HNO3 (dilute) ——–>
(b) Explain why
(i) H2O is a liquid while, inspite of a higher molecular mass, H2S is a gas.
(ii) Iron dissolves in HCl to form FeCl2 and not FeCl3.
(iii) Helium is used in diving equipment.
Answer:
(a) (i) P4 + 3NaOH + 3H20 ——-> PH3 + 3NaH2PO2
(ii) 3Cu + 8HN03(dilute) ——–> 3Cu(N03)2(aq) + 2NO(g) + 4H20(l)
(b) (i) It is because H20 is associated with intermolecular H-bonding, whereas H2S is not.
(ii) Refer Ans. to Q.20 (b) (i).
(iii) It is because helium does not dissolve as it is inert and lighter in blood therefore, does not cause bends or pain.

Question 165:
(a) Draw the structures of the following:
(i) N2O5 (ii) XeOF4
(b) Explain the following observations:
(i) The electron gain enthalpy of sulphur atom has greater negative value than that of oxygen atom.
(ii) Nitrogen does not form pentahalides.
(iii) In aqueous solution HI is a stronger acid than HCl.

Answer:
(a) (i) Refer Ans. to Q.40 (ii).
(ii) Refer Ans. to Q.13 (b) (ii).
(b) (i) Refer Ans. to Q.117 (i).
(ii) Refer Ans. to Q.17 (a) (iii).
(iii) It is because H—I has lower bond dissociation enthalpy thajn HCl due to longer bond length.

Question 166:
(a) Draw the structures of the following:
(i) H3PO2 (ii) BrF3
(b) How would you account for the following observations:
(i) Phosphorus has a greater tendency for catenation than nitrogen.
(ii) Bond dissociation energy of fluorine is less than that of chlorine.
(iii) No chemical compound of helium is known.

Answer:
(a) (i) Refer Ans. to Q.59 (ii).
(ii) Refer Ans. to Q.64 (b) (i).
(b) (i) Refer Ans. to Q.87 (b) (i).
(ii) Refer Ans. to Q. 18 (iii).
(iii) Refer Ans. to Q.81 (ii).

Question 167:
(a) Draw the structures of the following:
(i) PCl5(s)
(ii) S032-
(b) Explain the following observations:
(i) Ammonia has a higher boiling point than phosphine.
(ii) Helium does not form any chemical compound.
(iii) Bi(V) is a stronger oxidising agent than SB(V).
Answer:


(b) (i) Ammonia molecules are associated with intermolecular H-bonding. Whereas PH3 does not for H-bond.
(ii) Refer Ans. to Q.81 (ii).
(iii) Refer Ans. to Q. 15 (a) (i).

Question 168:
(a) Complete the following reaction equations:
(i) XeF2 + PF5 ——>
(ii) Cl2(g) + NaOH(aq)(hot & conc.) ———–>
(b) Explain the following observations:
(i) + 3 oxidation state becomes more and more stable from As to Bi in the group.
(ii) Sulphur in vapour state exhibits paramagnetism.
(iii) Fluorine does not exhibit any positive oxidation state.
Answer:
(a) (i) XeF2 + PF5 ———> [XeF]+ [PF6]-
(ii) Refer Ans. to Q.89 (a) (i).
(b) (i) It is due to inert pair effect.
(ii) Refer Ans. to Q.41 (i).
(iii) Refer Ans. to Q.65 (a) (ii).

Question 169:
(a) Complete the following reaction equations:
(i) SO2 + Mn04- + H20 ——->
(ii) HgCl2 + PH3 ———–>
(b) Explain the following observations:
(i) Sulphur has a greater tendency for catenation than oxygen.
(ii) Fluorine is a stronger oxidising agent than chlorine.
(iii) The +5 oxidation state becomes less stable down the group in group 15 of the periodic table.
Answer:
(a) (i) 5SO2 + 2MnO- + 2H2O ——–> 5SO4– + 2Mn2+ + 4H+
(ii) Refer Ans. to Q.83 (i).
(b) (i) Refer Ans. to Q.65 (a) (iii).
(ii) Refer Ans. to Q. 145 (b) (ii).
(iii) Refer Ans. to Q.86 (b) (ii).



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