
The geometries of the ammonia complexes of $N{i^{ + 2}}$ , $P{t^{ + 2}}$ and $Z{n^{ + 2}}$ respectively are:
A.Octahedral, square planar and tetrahedral
B.Square planer, octahedral and square planer
C.Tetrahedral, square planar and octahedral
D.Octahedral, tetrahedral and square planar
Answer
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Hint:Ammonia has 4 regions of electron density around the central nitrogen atom. These are arranged in a tetrahedral shape. The resulting molecular shape is trigonal pyramidal with \[H - N - H\] angles. $N{i^{ + 2}}$ , $P{t^{ + 2}}$ and $Z{n^{ + 2}}$ forms complexes with ammonia, they contain shape by their complexes. These complex forms form the metal and ligand.
Complete step by step answer:
Ammonia reacts with the metal ions and forms complexes. It reacts with $N{i^{ + 2}}$ , $P{t^{ + 2}}$ and $Z{n^{ + 2}}$ and form complexes of ${[Ni{(N{H_3})_6}]^{ + 2}}$ , ${[Pt{(N{H_3})_4}]^{ + 2}}$ and ${[Zn{(N{H_3})_4}]^{ + 2}}$ .
$N{i^{ + 2}}$ with $N{H_3}$ shows coordination number=6, it forms ${[Ni{(N{H_3})_6}]^{ + 2}}$ .
$P{t^{ + 2}}$ with $N{H_3}$ shows coordination number=4, which forms ${[Pt{(N{H_3})_4}]^{ + 2}}$
$Z{n^{ + 2}}$ with $N{H_3}$ shows coordination number=4, it forms ${[Zn{(N{H_3})_4}]^{ + 2}}$
Here, ${[Ni{(N{H_3})_6}]^{ + 2}}$ is an octahedral, ${[Pt{(N{H_3})_4}]^{ + 2}}$ is square planar and ${[Zn{(N{H_3})_4}]^{ + 2}}$ is tetrahedral in shape.
Hence, Option (A) is the correct answer.
Octahedral, square planar and tetrahedral.
Additional information:
The ammonia has a trigonal pyramidal shape with three hydrogen atoms and an unshared pair of electrons attached with nitrogen atoms. It is polar and associated because of strong intermolecular hydrogen bonding. The molecular geometry of $N{H_3}$ like its Lewis structure, electron geometry, hybridization, bond angles and molecular shape.
Note:Ammonia contains lone pair electrons. The geometry for a molecule with four electron pairs is tetrahedral, one of electron pairs is a lone pair rather than a bonding pair. The molecular geometry of $N{H_3}$ is called trigonal pyramidal. One or more of the molecular geometries. The domain geometry for a molecule with four electron pairs is tetrahedral, as was seen with $C{H_4}$ .
Complete step by step answer:
Ammonia reacts with the metal ions and forms complexes. It reacts with $N{i^{ + 2}}$ , $P{t^{ + 2}}$ and $Z{n^{ + 2}}$ and form complexes of ${[Ni{(N{H_3})_6}]^{ + 2}}$ , ${[Pt{(N{H_3})_4}]^{ + 2}}$ and ${[Zn{(N{H_3})_4}]^{ + 2}}$ .
$N{i^{ + 2}}$ with $N{H_3}$ shows coordination number=6, it forms ${[Ni{(N{H_3})_6}]^{ + 2}}$ .
$P{t^{ + 2}}$ with $N{H_3}$ shows coordination number=4, which forms ${[Pt{(N{H_3})_4}]^{ + 2}}$
$Z{n^{ + 2}}$ with $N{H_3}$ shows coordination number=4, it forms ${[Zn{(N{H_3})_4}]^{ + 2}}$
Here, ${[Ni{(N{H_3})_6}]^{ + 2}}$ is an octahedral, ${[Pt{(N{H_3})_4}]^{ + 2}}$ is square planar and ${[Zn{(N{H_3})_4}]^{ + 2}}$ is tetrahedral in shape.
Hence, Option (A) is the correct answer.
Octahedral, square planar and tetrahedral.
Additional information:
The ammonia has a trigonal pyramidal shape with three hydrogen atoms and an unshared pair of electrons attached with nitrogen atoms. It is polar and associated because of strong intermolecular hydrogen bonding. The molecular geometry of $N{H_3}$ like its Lewis structure, electron geometry, hybridization, bond angles and molecular shape.
Note:Ammonia contains lone pair electrons. The geometry for a molecule with four electron pairs is tetrahedral, one of electron pairs is a lone pair rather than a bonding pair. The molecular geometry of $N{H_3}$ is called trigonal pyramidal. One or more of the molecular geometries. The domain geometry for a molecule with four electron pairs is tetrahedral, as was seen with $C{H_4}$ .
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