Question 1

The  C-C  bond length is maximum in:

Accepted Answer

It diamond  C-C  bond, only  \sigma  bond is present while in case of graphite and fullerene  (C_{60}  and  C_{70})   C-C  bond contain doubled bond character. That is why diamond is having a maximum  C-C  bond length.

Question 2

Basic unit of zeolite, mica, asbestos and feldspar is:

Accepted Answer

Basic unit of zeolite, mica, asbestos and feldspar is  SiO_4^{4-} .

Question 3

{C}_{60}  an allotrope of carbon contains:

Accepted Answer

In  C_{60}  molecule there are  20  hexagons and  12  pentagons

Question 4

Match the items in Column I with its main use listed in Column II :Column IColumn IIASilica geliTransistorBSiliconiiIon-exchangerCSiliconeiiiDrying agentDSilicateivSealant

Accepted Answer

A. Silica gel is a good dehydrating agent. It is used as a dehydrating agent. It is used to absorb moisture.B. Silicon is used in transistors as it is a semiconductor.C. Silicon is a good resistant to heat and water. So it used as sealant to seal some element.D. Silicate including zeolites used in ion-exchange beds in domestic and commercial water purification, softening and other applications.

Question 5

The correct order of catenation is:

Accepted Answer

Solution:- (A)  C > Si > Ge \approx Sn As we move down the group, bond strength decreases, thereby decreasing the catenation tendency.Hence the order is as expected :  C > Si > Ge \approx  Sn

Question 6

The green colour produced in the borax bead test of a chromium (III) salt is due to:

Accepted Answer

In the borax bead test, metal oxide reacts with a glassy bead to form metaborates. Na_{2} B_{4} O_{7} . 10H_{2}O\xrightarrow {	riangle} Na_{2}B_{4}O_{7} \xrightarrow {	riangle} NaBO_{2} + B_{2}O_{3} Cr_{2}O_{3} + B_{2}O_{3} \xrightarrow {	riangle} \underset {Green}{Cr(BO_{2})_{3}} .Hence, option C is correct.

Question 7

Among the following, the correct statement(s) is (are):

Accepted Answer

The statements  (A), (B) and (D) are correct.(A)  BH_3  has the three-centre two-electron bonds in its dimeric structure as it is electron-deficient species. The three-centre two-electron bonds correspond to hydrogen bridges between 2 B atoms.(B)  Al(CH_3)_3  has the three-electron two-electron bonds in its dimeric structure as it is electron-deficient species. The three-centre two-electron bonds correspond to carbon bridges between 2 Al atoms.(D) The Lewis acidity of  BCl_3  is greater than that of  AlCl_3  as the bond formed with the base will involve 2p orbital overlap which is stronger than the 3p orbital overlap in the case of  AlCl_3 .The statement (C) is incorrect. AlCl_3  has the three-centre four-electron bonds in its dimeric structure. These bonds correspond to chlorine bridges between 2 Al atoms.

Question 8

The correct statement(s) for orthoboric acid is/are:

Accepted Answer

(A) It does not undergo self ionisation in water but accepts an electron pair from water, so it behaves as weak monobasic acid. Hence, (A) is incorrect. H_3BO_3 + H - OH  \rightleftharpoons B(OH)_4^- + H^+         ... (i)(B) When treated with 1, 2-dihydroxy or polyhydroxy compounds, they form chelate (ring complex) which effectively removes  [B(OH)_4^-]  species from solution and thereby produce a maximum number of  H_3O^+  or  H^+  ions, i.e. results in increased acidity.(C) Boric acid crystallises in a layer structure in which planar triangular  BO_3^{3-}  ions are bonded together through hydrogen bonds.(D) H _3 BO _3  is a weak electrolyte in water as in water the pKa value of  H_3BO_3  is 9.25.

Question 9

The crystalline form of borax has :

Accepted Answer

Correct formula of borax is  Na_2[B_4O_5(OH)_4].8H_2O .Borax has tetranuclear.  [B_4O_5(OH)_4]^{2-}  unit.Only two B atom lies in the same plane.Two Boron are  sp^2  & two are  sp^3  hybridised. One terminal hydroxide is present per boron atom.Hence, options  A ,  C  and  D  are correct.

Question 10

The increasing order of atomic radii of the following Group 13 elements is :

Accepted Answer

The increasing order of atomic radii of group 13 elements is : \displaystyle Ga  (135 \: pm) < Al   ( 143 \: pm) < In  ( 167 \: pm) < Tl  ( 170 \: pm) On moving down the group, the atomic radii increases. However due to poor shielding of nuclear charge by  3d  electrons,  \displaystyle \underset{(135pm)}{r_{Ga}} < \underset{(143pm)}{r_{Al}}   Ga  has 10 d electrons which do not screen the nuclear charge in an effective manner. The electrons in the outermost shell of  Ga  experience more force of attraction towards nucleus than  Al  which decreases atomic radius.Hence, the correct option is B.

	Column I		Column II
A	Silica gel	i	Transistor
B	Silicon	ii	Ion-exchanger
C	Silicone	iii	Drying agent
D	Silicate	iv	Sealant

Some P-Block Elements