The correct order of electronegativity of hybrid orbitals of carbon is:

1. sp > sp² < sp³             

2. sp > sp² > sp³

3. sp < sp² > sp³             

4. sp < sp² < sp³

${\mathrm{H}}_2\mathrm{O}$ has a net dipole moment while
\(BeF_2\) has zero dipole moment because :

1. F is more electronegative than oxygen.

2. Be is more electronegative than oxygen. 

3. ${\mathrm{H}}_2\mathrm{O}$ $$ molecule is linear and \(BeF_2\) is bent.

4. ${\mathrm{BeF}}_2$ molecule is linear and ${\mathrm{H}}_2\mathrm{O}$ is bent.

Which of the following compound have the same no. of lone pair on their central atom?

(a)${\mathrm{XeF}}_5^{- }$             (b) ${\mathrm{BrF}}_3$                 (c) ${\mathrm{XeF}}_2$                     (d) ${\mathrm{H}}_3{\mathrm{S}}^{+}$               

(e) Triple Methylene     

options are as follows :        

(a) (iv) and (v)                                     (b) (i) and (iii) 

(c) (i) and (ii)                                       (d) (ii) (iv), (v)

In which of the following species central atom is NOT surrounded by exactly 8 valence electrons?

(a) $B{F_4}^{-}$

(b) $NC{l}_3$

(c) PC${l_4}^{+}$

(d) $S{F}_4$

Which bond is expected to be the least polar?

(1) O-F

(2) P-F

(3) Si-N

(4) B-F 

Which set contains only covalently bonded molecules?

(a) $BC{l}_{3 }, SiC{l}_4 ,PC{l}_3$

(b) $N{H}_{4}Br , N_2{H}_4,HBr$

(c) $I_{2 ,}{H}_{2}S , NaI$

(d) $A{l}_2,O_{3 }, A{S}_4$

$N{F}_3$ is:

(1) non-polar compound 

(2) electrovalent compound 

(3) having low value of dipole moment than $N{H}_3$

(4) having more dipole moment than $N{H}_3$

Which of the following statement is true for $I{O}_2{F_2}^{-}$according to VSEPR theory?

(A) The lone pair and two I-O double bonds occupy the equatorial positions of trigonal
bipyramidal.

(B) It has $s{p}^3$ hybridization and is T-shaped.

(C)  Its structure is analogous to $S{F}_4$.

(D) (A) and (C) both

Which of the following is most covalent?

1. \(AlF_3\)

2. \(AlCl_3\)

3. \(AlBr_3\)

4. \(AlI_3\)

Among LiCl, BeCl₂, BCl₃ and CCl₄, the covalent bond character follows the order-

1. LiCl < BeCl₂ > BCl₃ > CCl₄

2. LiCl > BeCl₂ < BCl₃ < CCl₄

3. LiCl < BeCl₂ < BCl₃ < CCl₄

4. LiCl > BeCl₂ > BCl₃ > CCl₄