Question

Fill in the blanks.

ⅰ) $[CO(NO_2)_6]^{3-}$ the co-ordination number is ------ ii) Thiophene is ------ member heterocyclic compound iii) For resultant azimuthal quantum number L = 2 the term letter is ------ iv) ------ is an anthraquinone dye.

b) Select the correct alternative:

i) The hybridization in $[Zn(NH_3)_4]^{2+}$ a) $sp^3$ b) $dsp^2$ c) $sp^2$ d) None

ii) The ground term symbol for $d^6$ a) $^5D_4$ b) $^3F_2$ c) $^4F_{3/2}$ d) None

iii) In pyrrole each ring atom is in state of ------ hybridization. a) $sp$ b) $sp^2$ c) $sp^3$ d) $dsp^2$

iv) Which of the following IR active a) $H_2$ b) $N_2$ c) $CO$ d) $F_2$

• A. $sp^3$
• B. $dsp^2$
• C. $sp^2$
• D. None

Answer 1

Answer: (A)

Option (a) $sp^3$

Answer 2

Fill in the blanks:

i) In the complex $[Co(NO_2)_6]^{3-}$ the metal ion is coordinated to 6 nitrite ions. Answer: 6

ii) Thiophene is a five‐membered heterocyclic compound (with a sulfur atom in the ring). Answer: five‑membered

iii) For $L=2$, the corresponding term symbol letter is “d” (with $L=0,1,2,3, …$ corresponding to s, p, d, f,… respectively). Answer: d

iv) Alizarin is a well-known anthraquinone dye. Answer: Alizarin

Multiple Choice:

i) The hybridization in $[Zn(NH_3)_4]^{2+}$: Zn(II) has a $d^{10}$ configuration and forms a tetrahedral complex. The hybridization is $sp^3$. Answer: Option (a) $sp^3$

ii) The ground term symbol for a $d^6$ configuration: Using Hund’s rules, the maximum multiplicity for $d^6$ leads to a $^5D$ term. For a more than half-filled shell, the lowest $J$ value is highest, so $J = 4$. Answer: Option (a) $^5D_4$

iii) In pyrrole each ring atom is in the state of: In pyrrole, all ring atoms (both carbon and nitrogen) are $sp^2$ hybridized. Answer: Option (b) $sp^2$

iv) Which of the following is IR active: Homonuclear diatomics ($H_2$, $N_2$, $F_2$) do not show a dipole moment change, but CO, being heteronuclear, is IR active. Answer: Option (c) CO