NEET previous years Chemistry Q-A

 Class 11

Chapter 4 [Chemical bonding and Molecular structure]

1. Which one of the following sequences represents the correct increasing order of bond angle in the bond angles in the given molecules? 

(a) H2O <OF2<OCl2<ClO2

(b) OCl2<ClO2<H2O<OF2

(c) OF2<H2O<OCl2<ClO2

(d) ClO2<OF2<OCl2<H2O

Ans:- (c) 

Reason:- Water is sp3 hybridised with bond angle 104.5° due to the presence of two lone pairs. OF2 has structure similar to H2O with bond angle 103° due to higher Electronegativity of Fluorine. OCl2 also has similar structure with bond angle 111° because of steric crowding of two chloride atoms. However, ClO2 has π bond character with an odd electron so that bond angle is 118° . 

Class 12

Coordination compounds

1. Which complex compound obeys 18-electron (a) [V(CO)5]  (b) [Fe(NH3)6]2+

(c) [Ni(CO)6] (4) [Mn(H2O)6]2+

Ans:- (b)

Explanation:-(b) The complex which contains 18 valence electrons,follows 18-electron rule.

(a) [V(CO)5]: The number of valence electrons= 5+(2x5)=15e-

(b) [Fe(NH3)6]2+: The number of valence electrons = =6+(6x2)=6+12=18 e-

(c) [Ni(CO)6], : The number of valence electrons=10+(2 × 6)=22 e-

(d) [Mn(H2O)6]2+ : The number of valence electrons=5+(6x2)=17 e-

Thus, only [Fe(NH3)6] follows 18-electron rule.

2.The hybridization, oxidation number of central metal ion and shape of Wilkinson's catalyst are 

(a) dsp2, +1, square planar 

(b) sp3,.  +4, tetrahedral

(c) sp3d, +2, trigonal bipyramidal

(d) d2sp3, +6, octahedral 

Ans:-(a)

Explanation

In Wilkinson's catalyst-(a homogeneous catalyst), (Ph3P)3RhCI, Rh is dsp2 hybridised, has square planar shape and is in +1 oxidation state.

In complex [Rh(Ph3P)3],

 if x= oxidation state of Rh

x+0+(-1)=0

x=+1

3. The oxidation number of S in tetrathionate (S4O6^2-) is 

(a) +5 (b) 0 (c) 2.5 (d) all of these.