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Standard XII

Chemistry

Oxidation States

Ammonia acts ...

Question

Ammonia acts as a very good ligand but ammonium ion does not form complexes because:

N H_{3}

is a gas while

N H_{4}^{+}

in liquid form

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N H_{3}

undergoes

s p^{3}

hybridisation while

N H_{4}^{+}

undergoes

s p^{3} d

hybridisation

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N H_{4}^{+}

ion does not have any lone pair of electrons

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N H_{4}^{+}

ion has one unpaired electron while

N H_{3}

has two unpaired electrons.

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Solution

The correct option is C $N H_{4}^{+}$ ion does not have any lone pair of electrons
Complexes are formed by donation of a pair of electrons from ligand to metal. In ammonia, $N$ atom has one lone pair of electrons. So it can form complexes.
Nitrogen donates this lone pair of electrons to proton to form ammonium ion. $N H_{4}^{+}$ ion does not possess any lone pair of electrons which it can donate to central metal ion hence it does not form complexes.

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why nh3 is not an ion while nh4 is a ion?

Q. Assertion :Transition metal ion forming octahedral complexes undergo

s p^{3} d^{2}

d^{2} s p^{3}

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d^{2} s p^{3}

hybridisation whereas weak field ligands do not affect electronic configuration of the metal ion undergoes in

s p^{3} d^{2}

hybridisation.

Q. Which is/are not correct variation of indicated properties in

N H_{2}^{-}, N H_{3}

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N H_{4}^{+}

Q. Assertion :Cuprous ion (Cu

^{+}

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Q. The dissolution of ammonia gas in water does not obey Henry's law. On dissolving, a major portion o fammonia, molecules unite with

H_{2} O

to form

{N H}_{4} O H

molecules.

{N H}_{4} O H

again dissociates into

{N H}_{4}^{+}

and

{O H}^{-}

ions. In solution therefore, we have

{N H}_{3}

molecules,

{N H}_{4} O H

molecules and

{N H}_{4}^{+}

ions and the following equilibrium exist:

{N H}_{3} (g)

(pressure

P

and concentration

c

) Initially

⇌ {N H}_{3} (l) + H_{2} O ⇌ {N H}_{4} O H ⇌ {N H}_{4}^{+} + {O H}^{-}

Let

c_{1}

m o l / L

{N H}_{3}

pass in liquid state which on dissolution in water froms

c_{2}

m o l / L

{N H}_{4} O H

. The solution contains

c_{3}

m o l / L

{N H}_{4}^{+}

ions. The dissociation constant of

{N H}_{4} O H

can be given as:

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