
The INCORRECT statement is:
A.The spin - only magnetic moments of \[{{[Fe{{({{H}_{2}}O)}_{6}}]}^{2+}}\] and \[{{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}}\] are nearly similar
B.The spin-only magnetic moment of \[{{[Ni{{\left( N{{H}_{3}} \right)}_{4}}{{\left( {{H}_{2}}O \right)}_{2}}]}^{2+}}\]is 2.83 BM
C.The gemstone, ruby, has $C{{r}^{3+}}$ ions occupying the octahedral sites of beryl
D.The color of \[{{[COCl{{\left( N{{H}_{3}} \right)}_{5}}]}^{2+}}\] is violet as it absorbs the yellow light
Answer
552.6k+ views
Hint:To solve this type of question we have to check every statement and we should have prior knowledge about the terms mentioned in the question. The spin only magnetic moment is the moment which is caused due to the spin of the particle. It can be found by writing the correct electronic configuration.
Complete answer:
Let’s check every statement:
- The spin - only magnetic moments of \[{{[Fe{{({{H}_{2}}O)}_{6}}]}^{2+}}\] and \[{{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}}\]are nearly similar
Spin only magnetic moment is calculated by using the formulae:
\[\mu =\sqrt{n(n+2)}B.M.\]
Where n is the number of unpaired electrons
Spin only magnetic moment of $F{{e}^{+2}}$
Number of unpaired electrons = 4
So, \[\mu =\sqrt{n(n+2)}B.M.\] = $\sqrt{4(4+2)}$= 4.899 B.M.
Spin only magnetic moment of $C{{r}^{+2}}$
Number of unpaired electrons = 4
So, \[\mu =\sqrt{n(n+2)}B.M.\] = $\sqrt{4(4+2)}$= 4.899 B.M.
This statement is correct
-The spin-only magnetic moment of \[{{[Ni{{\left( N{{H}_{3}} \right)}_{4}}{{\left( {{H}_{2}}O \right)}_{2}}]}^{2+}}\]is 2.83 BM
Spin only magnetic moment of $N{{i}^{+2}}$
Number of unpaired electrons = 2
So, \[\mu =\sqrt{n(n+2)}B.M.\] = $\sqrt{2(2+2)}$= 2.83 B.M.
This statement is correct
-The gemstone, ruby, has $C{{r}^{3+}}$ ions occupying the octahedral sites of beryl
Ruby has $C{{r}^{3+}}$ ions occupying the octahedral sites of aluminium oxides normally occupied by 3+ ion of aluminium.
This statement is incorrect.
-The color of \[{{[COCl{{\left( N{{H}_{3}} \right)}_{5}}]}^{2+}}\]is violet as it absorbs the yellow light
This statement is correct.
Hence the correct answer is option (C) .
Note:
The number of unpaired electrons are determined using the electronic configuration and while writing the electronic configuration we should take care of the charge present on the element. If there is a positive charge on the element remove that number of electrons while writing the electronic configuration similarly if there is a negative charge on the element then pass that number of electron while writing the electronic configuration.
Complete answer:
Let’s check every statement:
- The spin - only magnetic moments of \[{{[Fe{{({{H}_{2}}O)}_{6}}]}^{2+}}\] and \[{{[Cr{{({{H}_{2}}O)}_{6}}]}^{2+}}\]are nearly similar
Spin only magnetic moment is calculated by using the formulae:
\[\mu =\sqrt{n(n+2)}B.M.\]
Where n is the number of unpaired electrons
Spin only magnetic moment of $F{{e}^{+2}}$
Number of unpaired electrons = 4
So, \[\mu =\sqrt{n(n+2)}B.M.\] = $\sqrt{4(4+2)}$= 4.899 B.M.
Spin only magnetic moment of $C{{r}^{+2}}$
Number of unpaired electrons = 4
So, \[\mu =\sqrt{n(n+2)}B.M.\] = $\sqrt{4(4+2)}$= 4.899 B.M.
This statement is correct
-The spin-only magnetic moment of \[{{[Ni{{\left( N{{H}_{3}} \right)}_{4}}{{\left( {{H}_{2}}O \right)}_{2}}]}^{2+}}\]is 2.83 BM
Spin only magnetic moment of $N{{i}^{+2}}$
Number of unpaired electrons = 2
So, \[\mu =\sqrt{n(n+2)}B.M.\] = $\sqrt{2(2+2)}$= 2.83 B.M.
This statement is correct
-The gemstone, ruby, has $C{{r}^{3+}}$ ions occupying the octahedral sites of beryl
Ruby has $C{{r}^{3+}}$ ions occupying the octahedral sites of aluminium oxides normally occupied by 3+ ion of aluminium.
This statement is incorrect.
-The color of \[{{[COCl{{\left( N{{H}_{3}} \right)}_{5}}]}^{2+}}\]is violet as it absorbs the yellow light
This statement is correct.
Hence the correct answer is option (C) .
Note:
The number of unpaired electrons are determined using the electronic configuration and while writing the electronic configuration we should take care of the charge present on the element. If there is a positive charge on the element remove that number of electrons while writing the electronic configuration similarly if there is a negative charge on the element then pass that number of electron while writing the electronic configuration.
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