Question 1

Single  N-N  bond is weaker than the single  P-P  bond. This is because of:

Accepted Answer

N-N  single bond is weaker than  P-P  bond due to smaller size of  N  as compared to  P . Smaller size of  N  leads to smaller  N-N  bond length. As a result, the lone pair of electrons on the both the  N  atoms repel each other leading to unstability or weakening of  N-N  bond. Because of larger size of  P  atom,  P-P  bond length is more and lone pair-lone pair repulsion between  P  atoms is less which makes the  P-P  bond stronger than  N-N  bond.

Question 2

Anomalous behaviour of oxygen, compared to other VIA group elements is due to:

Accepted Answer

Oxygen is the first element of VI A group and shows Anaomalous behaviour than other group elements because1:High Electronegativity2:small atomic size3:Non-avalability of d-orbits results in different compounds than their other group elements.

Question 3

Explain why:Hydride of sulphur is a gas while hydride of oxygen is a liquid.

Accepted Answer

Hydride of sulphur is  H_2S  and that of oxygen is  H_2O . Since oxygen being highly electronegative has a tendency to form hydrogen bond, thereby increases the intermolecular interaction through extensive H-bonding. While,  S  can not undergo H-bonding and this explains the liquid phase of water and gaseous phase of  H_2S .

Question 4

Given reasons for the following:Dioxygen is a gas but sulphur a solid.

Accepted Answer

Oxygen is smaller in size as compared to sulphur due to which, it can effectively form  p \pi - p \pi  bonds and form  {O}_{2}  molecule. Also, the intermolecular forces in oxygen are weak van der Waal&#8217;s, which cause it to exist as a gas. On the other hand, sulphur does not form  {S}_{2}  molecule but exists as a puckered structure held together by strong covalent bonds. Hence, it is solid.

Question 5

Which of the following halogens doesn't exhibit positive oxidation state in its compounds?

Accepted Answer

Fluorine, the most electronegative element, has no positive oxidation states, but the other halogens include chlorine, bromine, and iodine commonly exhibit +1, +3, +5, and +7 states.Hence, option  D  is correct.

Question 6

In which one of the following compounds does sulphur have the least oxidation number?

Accepted Answer

A:  SO_{2}\rightarrow S^{+4} B:  H_{2}SO_{4}\rightarrow S^{+6} C:  Na_{2}S_{2}O_{3}\rightarrow S^{+2},S^{+2} D:  Na_{2}S_{4}O_{6}\rightarrow S^{0},S^{+5}

Question 7

Which one of the following acid possesses oxidising, reducing and complex forming properties?

Accepted Answer

In  HNO_2 , N is in the intemediate oxidation state between its highest and lowest possible oxidation state so it can act as oxidising as well as reducing agent. Also, it forms  NO_2^-  ion which forms complex.

Question 8

Assertion (A): Bismuth compounds are stable in +V oxidation state than in +III oxidation state.Reason (R): Inert pair effect is observed in Bismuth compounds.

Accepted Answer

Inert pair effect is observed in Bismuth compoundsSo +III state is more stable than +V oxidation state

Question 9

Group 16 elements have a lower value of first ionization enthalpy as compared to group 15 elements because:

Accepted Answer

Group 16 elements have a lower value of first ionization enthalpy as compared to group 15 elements. As group 15 elements have half filled p-orbital due to which group 15 have got extra stability.

Question 10

Question 11

Bond dissociation enthalpy of EH (E = element) bonds is given below. Which of the compounds will act as strongest reducing agent?Compound NH_{3} PH_{3} AsH_{3} SbH_{3} \Delta_{diss}   (E-H)/kJmol^{-1} 389322297255

Accepted Answer

As the bond dissociation energy decreases reducing nature of the substance increases because the bond can be broken down easily and can donate electron easily and make the other substance to be reduced.So the bond dissociation energy is inversely proportional to reducing nature.The order of bond dissociation energy of given molecules is  NH_3>PH_3>AsH_3>SbH_3 due to increase in size of the central atom.So the strongest reducing agent is  SbH_3 .Hence option D is correct.

Question 12

Which of the following orders regarding thermal stability of hydrides  MH_3  of group  15  is correct?

Accepted Answer

correct order of thermal stability is  NH_3 > PH_3 > AsH_3

Question 13

The oxyacid of phosphorus that is reducing, dibasic and contains one  P - O - P  bond is :

Accepted Answer

H_3PO_3  is reducing, dibasic and has no  P-O-P  bond. H_4P_2O_7  is tetrabasic, has one  P-O-P  bond and is reducing. H_3PO_4  has no  P-O-P  bond, is tribasic and non-reducing. H_4P_2O_5  is dibasic, has  P-O-P  bond and is reducing.

Question 14

In which of the following option (s) all species contains X-O-X bond(s) in structure ( X =  central atom)?

Accepted Answer

From all the oxides those have  X-O-X  bonds(s) [X=Central atom] are  P_4O_{10},P_4O_6 H_5P_3O_{10},S_3O_9,Si_2O^{6-}_7,N_2O_5 Refer image,From all the above data and pictures option Bi.e  P_4O_{10},P_4O_6,H_5P_3O_{10}  and Option D  i.e  S_3O_9,P_4O_6,Si_2O^{6-}_7  is correct.

Question 15

The weakest acid among the following is :

Accepted Answer

Hint: The acidic strength of a species depends on its  H^+ &#10;ion releasing tendency.&#10;&#10;Explanation:As the hydrogen ion releasing tendency increases acidity of the species also increases.&#10;&#10;In  H_2SO_4  , S  has oxidation state  {+2}\ +\ x\ +\ {(-8)}=0 x=+6 .where  x  is the oxidation state of  S .&#10;&#10;In  H_2SO_3   S  has oxidation state  {+2}\ +\ {x}\ +\ {(-6)}=0 x=+4 &#10;&#10;As in  H_2SO_4 ,  S  has higher oxidation state&#10;so it will attracts the electron pair strongly. This will easily remove  H^+ &#10;ion and stabilize the conjugate base.&#10;&#10;In case of  H_2SeO_3  and  H_2SeO_4   Se  have&#10;a larger size than  S .&#10;&#10;So, it attracts electron pairs not that strong as  S  does. So it can&#8217;t release  H^+  ion easily.&#10;&#10; Se  in  H_2SeO_3  has low oxidation state&#10; +4  than in  H_2SeO_4  which is  +6 .&#10;&#10;So,  H_2SeO_3  will not release hydrogen ions easily.&#10;&#10;Hence,  H_2SeO_3  is the weakest acid.&#10;&#10;Final Answer: Correct option  (D) .

Question 16

Which among the following is the most reactive?

Accepted Answer

\bf{Hint:}  A less stable compound or molecule will be more reactive. \bf{Explanation:} The reactivity of any molecule or compound depends on its stability.A stable molecule does not react with other molecules until unless an external force like temperature, highly reactive reagent is not applied to it.In given molecules,  ICl  is the least stable molecule due to the large size difference as well as electronegativity difference between both atoms present in this molecule which makes it highly reactive.On other hand,  Cl_{2} ,  Br_{2}  and  I_{2}  are stable molecules at room temperature which makes these less reactive than  ICl .Also in the case of  Cl_2 ,  Br_2  and  I_2  there is no electronegativity difference between the element so the molecule is non-polar so they are least reactive compared to  ICl \bf{Final \ answer:}  Option  D

Question 17

(i) Why are interhalogen compounds more reactive than halogens ?(ii) All the five bonds in  PCl_5  are not equivalent. Justify.

Accepted Answer

(i) Interhalogen compounds are more reactive than all halogens except  F . The interhalogen bonds (A-X) are weaker than the bonds between halogen atoms (X-X) in dihalogen compounds. The main reason for this is that the electronegativities of atoms in a diahalogen compound are the same while that of atoms in interhalogen compounds are different. Thus, the bond in interhalogen compounds is polarised, because of which it can break more easily. For example, the bond between  ICl  is weaker than  I_2  because  ICl  atom is more polarised due to a difference in electronegativities of the two atoms.(ii) The five bonds in  PCl_5  are not equivalent.  PCl_5  is  sp^3d  hybridised. Two chlorine atoms are located axially at the top and bottom of the molecule while the remaining three are attached equatorially. The length of the axial and equatorial bonds between  P  and  Cl  is different.

Question 18

Identify the incorrect match related to structure of given interhalogen compounds.

Accepted Answer

In the structure of  ClF_5 Due to both elements having 7 valence electrons and forming covalent bonds, it makes sense that chlorine forms 5 covalent bonds with the 5 fluorine molecules, which would normally form a trigonal bipyramidal shape. However, this only uses up 5 of chlorine&#8217;s 7 electrons, a further two are still unbonded. These form a lone pair, which has similar repulsion to an atom.Accounting for the lone pair alongside the 5 fluorines, we see that the shape formed is square pyramidal.So, option B is correct.

The P-Block Elements