Match the species in Column I with the 5 types of hybrid orbitals given  in Column II:

	Column I		Column II
1	\(SF_4\)	(a)	\(sp^3d^2\)
2	\(IF_5\)	(b)	\(d^2sp^3\)
3	\(NO^+_2\)	(c)	\(sp^3d\)
4	\(NH^+_4\)	(d)	\(sp^3\)
		(e)	\(sp\)

1. (1) → (a), (2) → (b), (3) → (e), (4) → (d)

2. (1) → (c), (2) → (a), (3) → (e), (4) → (d)

3. (1) → (d), (2) → (a), (3) → (e), (4) → (c)

4. (1) → (e), (2) → (b), (3) → (c), (4) → (d)

Match the species in Column I with the shape in Column II.

1. (1) → (d), (2) → (a), (3) → (b), (4) → (c)

2. (1) → (b), (2) → (a), (3) → (e), (4) → (c)

3. (1) → (e), (2) → (a), (3) → (b), (4) → (c)

4. (1) → (c), (2) → (a), (3) → (b), (4) → (d)

Which of the following species absorb maximum energy in its HOMO-LUMO electronic tranistion ?

$1. {\mathrm{O}}_2$
$2. {\mathrm{N}}_2^{-}$
$3. {\mathrm{C}}_2$
$4. {\mathrm{N}}_2$

Which of the following facts given is not correct/

(I) Bond length order : ${\mathrm{H}}_2^{-} = {\mathrm{H}}_2^{+} > {\mathrm{H}}_2$

(II) ${\mathrm{O}}_2^{+}, \mathrm{NO}, {\mathrm{N}}_2^{-}$ have same bond order of $2\frac{1}{2}$

(III) Bond order can assume any value including zero upto four

(IV) ${\mathrm{NO}}_3^{-}$ and ${\mathrm{BO}}_3^{3-}$ have same bond order for X - O bond (where X is central atom)

1. I, II & III

2. I & IV

3. II & IV

4. I & II

${\mathrm{N}}_2$ and ${\mathrm{O}}_2$ are converted to monocations ${\mathrm{N}}_2^{+}$ and ${\mathrm{O}}_2^{+}$ respectively, which is wrong statements:

1. In ${\mathrm{N}}_2^{+}$, the $\mathrm{N}-\mathrm{N}$ bond weakens

2. In ${\mathrm{O}}_2^{+}$, the $\mathrm{O}-\mathrm{O}$ bond order increases

3. In ${\mathrm{O}}_2^{+}$, the paramagnetism decreases

4. ${\mathrm{N}}_2^{+}$ becomes diamagnetic

In which of the following transformations, the bond order has increased and the magnetic behavior has changed?

$1. {\mathrm{C}}_2^{+} \to {\mathrm{C}}_2$
$2. {\mathrm{NO}}^{+}\to \mathrm{NO}$
$3. {\mathrm{O}}_2 \to {\mathrm{O}}_2^{+}$
$4. {\mathrm{N}}_2 \to {\mathrm{N}}_2^{+}$

${\mathrm{BF}}_3$ and ${\mathrm{NF}}_3$ both are covalent compounds but ${\mathrm{NF}}_3$ is polar whereas ${\mathrm{BF}}_3$ is non-polar. This is because :

1. Nitrogen atom is smaller than boron atom

2. N-F bond is more polar than B-F bond

3. NF${}_3$ is pyramidal whereas BF${}_3$ is planar triangular

4. BF${}_3$ is electron deficient whereas NF${}_3$ is not

How many sp${}^2$ and sp-hybridised carbon atoms are present respectively in the following compound ?

1. 4, 2

2. 6, 0

3. 3, 3

4. 5, 1

Which molecule does not exist ?

$1. {\mathrm{OF}}_2$
$2. {\mathrm{OF}}_4$
$3. {\mathrm{SF}}_2$
$4. {\mathrm{SF}}_4$

The correct order of Cl-O bond order is

$1. {\mathrm{ClO}}_3^{-} < {\mathrm{ClO}}_4^{-} < {\mathrm{ClO}}_2^{-} < {\mathrm{ClO}}^{-}$
$2. {\mathrm{ClO}}^{-} < {\mathrm{ClO}}_4^{-} < {\mathrm{ClO}}_3^{-} < {\mathrm{ClO}}_2^{-}$
$3. {\mathrm{ClO}}^{-} < {\mathrm{ClO}}_2^{-} < {\mathrm{ClO}}_3^{-}< {\mathrm{ClO}}_4^{-}$
$4. {\mathrm{ClO}}_4^{-} < {\mathrm{ClO}}_3^{-} < {\mathrm{ClO}}_2^{-} < {\mathrm{ClO}}^{-}$