
In which of the following complex oxidation states of the metal is zero?
(A) $\left [ pt\left ( NH_3 \right )_2 Cl_2 \right ]$
(B) $\left [ Cr\left ( CO \right )_6 \right ]$
(C) $\left [ Cr\left ( NH_3\right )_3 Cl_3 \right ]$
(D) $\left [ Cr\left ( en\right )_2 Cl_2 \right ]$
Answer
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Hint: To calculate the oxidation state of the metal, just assume the oxidation state of the central metal, say x and add to it the oxidation state of ligands (multiply by the number of that ligand present). As in the given options, there is no anionic part thus equating the addition of the oxidation state of metal and ligands to zero.
Complete Step by Step Solution:
For \[\left[ pt{{\left( N{{H}_{3}} \right)}_{2}}C{{l}_{2}} \right]\], assume the oxidation state of pt (platinum), central metal ion, say x. Charge on ammonia, NH3 is zero as ammonia is a neutral ligand and charge on chlorine is -1 then
x + 2(0) + 2(-1) = 0
x – 2 = 0
x = +2
For \[\left[ Cr{{\left( CO \right)}_{6}} \right]\], the formal charge on CO is zero and let oxidation number of Cr (chromium), central metal say x such as
x + 6(0) = 0
x = 0
For \[\left[ Cr{{\left( N{{H}_{3}} \right)}_{3}}C{{l}_{3}} \right]\], ammonia is a neutral ligand, and the charge on the chlorine ligand is -1. Let oxidation state of Cr, central metal say x such as
x + 3(0) + 3(-1) = 0
x – 3 = 0
x = +3
For \[\left[ Cr{{\left( N{{H}_{3}} \right)}_{3}}C{{l}_{3}} \right]\], en(ethylenediamine) is a neutral molecule and hence no charge, and the charge on chlorine is -1. Let oxidation state of Cr, central metal say x such as
x + 2(0) + 2(-1) = 0
x – 2 = 0
x = +2
Thus, the correct option is B.
Note: Charge on CO is 0 as carbon is tetrahedral and can form four covalent bonds. But in CO carbon forms three bonds with oxygen because of this, carbon is left with one electron or can say attain charge -1 whereas oxygen loses one electron of its lone pair in forming the third bond thus oxygen attains +1. Both -1 and +1 will cancel out each other and the molecule, CO, becomes a neutral molecule.
Complete Step by Step Solution:
For \[\left[ pt{{\left( N{{H}_{3}} \right)}_{2}}C{{l}_{2}} \right]\], assume the oxidation state of pt (platinum), central metal ion, say x. Charge on ammonia, NH3 is zero as ammonia is a neutral ligand and charge on chlorine is -1 then
x + 2(0) + 2(-1) = 0
x – 2 = 0
x = +2
For \[\left[ Cr{{\left( CO \right)}_{6}} \right]\], the formal charge on CO is zero and let oxidation number of Cr (chromium), central metal say x such as
x + 6(0) = 0
x = 0
For \[\left[ Cr{{\left( N{{H}_{3}} \right)}_{3}}C{{l}_{3}} \right]\], ammonia is a neutral ligand, and the charge on the chlorine ligand is -1. Let oxidation state of Cr, central metal say x such as
x + 3(0) + 3(-1) = 0
x – 3 = 0
x = +3
For \[\left[ Cr{{\left( N{{H}_{3}} \right)}_{3}}C{{l}_{3}} \right]\], en(ethylenediamine) is a neutral molecule and hence no charge, and the charge on chlorine is -1. Let oxidation state of Cr, central metal say x such as
x + 2(0) + 2(-1) = 0
x – 2 = 0
x = +2
Thus, the correct option is B.
Note: Charge on CO is 0 as carbon is tetrahedral and can form four covalent bonds. But in CO carbon forms three bonds with oxygen because of this, carbon is left with one electron or can say attain charge -1 whereas oxygen loses one electron of its lone pair in forming the third bond thus oxygen attains +1. Both -1 and +1 will cancel out each other and the molecule, CO, becomes a neutral molecule.
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