Answer
Verified
386.4k+ views
Hint: If the compound has unpaired electrons then the compound is paramagnetic and if it has all the electrons in pairs then it is diamagnetic. In a complex compound, the diamagnetic nature is calculated by finding the oxidation number of the central metal atom and then its configuration.
Complete answer:
Let us study all the complexes one by one:
(i)- $[Co{{(N{{H}_{3}})}_{6}}]C{{l}_{3}}$
First, we have to calculate the oxidation number of cobalt ions.
The oxidation number of $N{{H}_{3}}$ is zero because it is a neutral ligand and the oxidation number of $Cl$ is -1. So, the oxidation number of cobalt is:
$x\text{ + 6(0) + 3(-1) = 0}$
$x\text{ = +3}$
So, the oxidation number of cobalt is +3.
The ground state electronic configuration of cobalt (27) is $4{{s}^{2}}3{{d}^{7}}$ and in $C{{o}^{3+}}$ state $4{{s}^{0}}3{{d}^{6}}$
It has 6 electrons which are paired.
Hence, $[Co{{(N{{H}_{3}})}_{6}}]C{{l}_{3}}$is diamagnetic.
(ii)- $[Ni{{(N{{H}_{3}})}_{6}}]C{{l}_{2}}$
First, we have to calculate the oxidation number of nickel ions.
The oxidation number of $N{{H}_{3}}$ is zero because it is a neutral ligand and the oxidation number of $Cl$ is -1. So, the oxidation number of nickel is:
$x\text{ + 6(0) + 2(-1) = 0}$
$x\text{ = +2}$
So, the oxidation number of nickel is +2.
The ground state electronic configuration of nickel (28) is $4{{s}^{2}}3{{d}^{8}}$ and in $N{{i}^{2+}}$ state $4{{s}^{0}}3{{d}^{8}}$
It has 2 unpaired electrons.
Hence, $[Ni{{(N{{H}_{3}})}_{6}}]C{{l}_{2}}$is paramagnetic.
(iii)- $[Cr{{({{H}_{2}}O)}_{6}}]C{{l}_{3}}$
First, we have to calculate the oxidation number of chromium ion.
The oxidation number of ${{H}_{2}}O$ is zero because it is a neutral ligand and the oxidation number of $Cl$ is -1. So, the oxidation number of chromium is:
$x\text{ + 6(0) + 3(-1) = 0}$
$x\text{ = +3}$
So, the oxidation number of chromium is +3.
The ground state electronic configuration of chromium (24) is $4{{s}^{1}}3{{d}^{5}}$ and in $C{{r}^{3+}}$ state $4{{s}^{0}}3{{d}^{3}}$
It has 3 unpaired electrons.
Hence, $[Cr{{({{H}_{2}}O)}_{6}}]C{{l}_{3}}$ is paramagnetic.
(iv)- $[Fe{{({{H}_{2}}O)}_{6}}]C{{l}_{2}}$
First, we have to calculate the oxidation number of iron ions.
The oxidation number of ${{H}_{2}}O$ is zero because it is a neutral ligand and the oxidation number of $Cl$ is -1. So, the oxidation number of iron is:
$x\text{ + 6(0) + 2(-1) = 0}$
$x\text{ = +2}$
So, the oxidation number of iron is +2.
The ground state electronic configuration of iron (26) is $4{{s}^{2}}3{{d}^{6}}$ and in $F{{e}^{2+}}$ state $4{{s}^{0}}3{{d}^{6}}$
It has 4 unpaired electrons.
Hence, $[Fe{{({{H}_{2}}O)}_{6}}]C{{l}_{2}}$ is paramagnetic.
So, the correct option is (a)- (i)
Note:
If the complex compound has a strong field ligand like $N{{O}_{2}}^{-}, C{{N}^{-}}, CO$, etc they will pair up the unpaired electrons and if the compound has a weak field ligand ${{H}_{2}}O, N{{H}_{3}},{{F}^{-}}$, etc, they will not pair up the unpaired electrons.
Complete answer:
Let us study all the complexes one by one:
(i)- $[Co{{(N{{H}_{3}})}_{6}}]C{{l}_{3}}$
First, we have to calculate the oxidation number of cobalt ions.
The oxidation number of $N{{H}_{3}}$ is zero because it is a neutral ligand and the oxidation number of $Cl$ is -1. So, the oxidation number of cobalt is:
$x\text{ + 6(0) + 3(-1) = 0}$
$x\text{ = +3}$
So, the oxidation number of cobalt is +3.
The ground state electronic configuration of cobalt (27) is $4{{s}^{2}}3{{d}^{7}}$ and in $C{{o}^{3+}}$ state $4{{s}^{0}}3{{d}^{6}}$
It has 6 electrons which are paired.
Hence, $[Co{{(N{{H}_{3}})}_{6}}]C{{l}_{3}}$is diamagnetic.
(ii)- $[Ni{{(N{{H}_{3}})}_{6}}]C{{l}_{2}}$
First, we have to calculate the oxidation number of nickel ions.
The oxidation number of $N{{H}_{3}}$ is zero because it is a neutral ligand and the oxidation number of $Cl$ is -1. So, the oxidation number of nickel is:
$x\text{ + 6(0) + 2(-1) = 0}$
$x\text{ = +2}$
So, the oxidation number of nickel is +2.
The ground state electronic configuration of nickel (28) is $4{{s}^{2}}3{{d}^{8}}$ and in $N{{i}^{2+}}$ state $4{{s}^{0}}3{{d}^{8}}$
It has 2 unpaired electrons.
Hence, $[Ni{{(N{{H}_{3}})}_{6}}]C{{l}_{2}}$is paramagnetic.
(iii)- $[Cr{{({{H}_{2}}O)}_{6}}]C{{l}_{3}}$
First, we have to calculate the oxidation number of chromium ion.
The oxidation number of ${{H}_{2}}O$ is zero because it is a neutral ligand and the oxidation number of $Cl$ is -1. So, the oxidation number of chromium is:
$x\text{ + 6(0) + 3(-1) = 0}$
$x\text{ = +3}$
So, the oxidation number of chromium is +3.
The ground state electronic configuration of chromium (24) is $4{{s}^{1}}3{{d}^{5}}$ and in $C{{r}^{3+}}$ state $4{{s}^{0}}3{{d}^{3}}$
It has 3 unpaired electrons.
Hence, $[Cr{{({{H}_{2}}O)}_{6}}]C{{l}_{3}}$ is paramagnetic.
(iv)- $[Fe{{({{H}_{2}}O)}_{6}}]C{{l}_{2}}$
First, we have to calculate the oxidation number of iron ions.
The oxidation number of ${{H}_{2}}O$ is zero because it is a neutral ligand and the oxidation number of $Cl$ is -1. So, the oxidation number of iron is:
$x\text{ + 6(0) + 2(-1) = 0}$
$x\text{ = +2}$
So, the oxidation number of iron is +2.
The ground state electronic configuration of iron (26) is $4{{s}^{2}}3{{d}^{6}}$ and in $F{{e}^{2+}}$ state $4{{s}^{0}}3{{d}^{6}}$
It has 4 unpaired electrons.
Hence, $[Fe{{({{H}_{2}}O)}_{6}}]C{{l}_{2}}$ is paramagnetic.
So, the correct option is (a)- (i)
Note:
If the complex compound has a strong field ligand like $N{{O}_{2}}^{-}, C{{N}^{-}}, CO$, etc they will pair up the unpaired electrons and if the compound has a weak field ligand ${{H}_{2}}O, N{{H}_{3}},{{F}^{-}}$, etc, they will not pair up the unpaired electrons.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE