MolSynapsequestion bank

MolSynapse / Question bank / Basic Inorganic Nomenclature and Oxidation Number / Oxidation number determination including fractional and unusual cases

JEE Main levelOne or more correct optionsDifficulty 2 of 3bn022

Select the species in which the AVERAGE oxidation number of carbon is zero (H is $+ 1$ , O is $- 2$ , Cl is $- 1$ ; in $CaC X_{2}$ , Ca is $+ 2$ ).

A $CH X_{2} Cl X_{2}$
B $CaC X_{2}$
C $CH X_{3} COOH$
D $C X_{6} H X_{12} O X_{6}$

Answer: A, C, D

C X_{6} H X_{12} O X_{6}

:

6 x + 12 - 12 = 0

gives

0

.

CH X_{3} COOH

:

2 x + 4 - 4 = 0

gives

0

.

CH X_{2} Cl X_{2}

:

x + 2 - 2 = 0

gives

0

.

CaC X_{2}

:

2 + 2 x = 0

gives

- 1

.

Practice Basic Inorganic Nomenclature and Oxidation Number More on Oxidation number determination including fractional and unusual cases

Previous questionNext question

More questions on Oxidation number determination including fractional and unusual cases