Question

When lead nitrate is strongly heated, the oxidation number of which of the following atoms change?
A) Only in Pb and N
B) Only in N and O
C) Pb, N and O
D) Only in N

Answer 1

Hint: The oxidation number of the elements is the oxidation state of the element. Or it can be said as the number of electrons an atom can share to form bonds.
Atomic number of Pb is 82, N is 7 and O is 8

Complete step by step solution:
The oxidation number is the number of electrons that an atom shared to form a bond or to involve in chemical reactions. Oxidation state and oxidation number almost remain same in most cases, but while considering the $Pb=\left[ Xe \right]4{{f}^{14}}5{{d}^{10}}6{{s}^{2}}6{{p}^{2}}$electronegativity of the atoms it gives a different meaning.
Lead nitrate when heated it forms lead oxide, nitrogen dioxide and oxygen. And the reaction can be written as,
\[2Pb{{\left( N{{O}_{3}} \right)}_{2}}\xrightarrow{\Delta }2PbO+4N{{O}_{2}}+{{O}_{2}}\]\[2Pb{{\left( N{{O}_{3}} \right)}_{2}}\xrightarrow{\Delta }2PbO+4N{{O}_{2}}+{{O}_{2}}\]
-The atomic number of Pb is 84 and has the electronic configuration as$Pb=\left[ Xe \right]4{{f}^{14}}5{{d}^{10}}6{{s}^{2}}6{{p}^{2}}$, and Pb has the common oxidation states +4 and +2.
In\[Pb{{\left( N{{O}_{3}} \right)}_{2}}\], Pb is having oxidation state +2, when it is converted into PbO , the oxidation state remains the same i.e. +2.
This can be calculated as, 1) \[Pb{{\left( N{{O}_{3}} \right)}_{2}}\to PbO\]
Here we know Pb has+2 oxidation state in lead nitrate and if we calculate the oxidation state of Pb in PbO i.e. we know for O oxidation number is -2 ,
Therefore x+ (-2) =0
X=2, here we gave x for the oxidation number of Pb and oxidation number for a molecule is zero.
So there is no change in oxidation number for Pb.
-For nitrogen: Atomic number is 7 and electronic configuration is$N=1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}$, oxidation number is +5
Here in lead nitrate, N is in +5 oxidation state when it is converted into nitrogen dioxide ($N{{O}_{2}}$)
$x+2(-2)=0$
\[x=4\]
So N has a change of oxidation state, $5\to 4$
-For Oxygen: Atomic number is 8 and has the electronic configuration $O=1{{s}^{2}}2{{s}^{2}}2{{p}^{4}}$$O=1{{s}^{2}}2{{s}^{2}}2{{p}^{4}}$
Since it can take up two electrons to obtain stable octet configuration it has -2 oxidation state.
-In lead nitrate, O is in -2 oxidation state and it is converted into an oxygen molecule which possesses the oxidation number zero.
So in O there is a change of oxidation number,$-2\to 0$ $-2\to 0$
So here except Pb others are undergoing change in oxidation number.

The correct option for the question is option (B).

Note: The oxidation states with their respective sign should be known to find out the oxidation state of the unknown atoms. Even though in lead nitrate ,if we don’t know oxidation state of Pb if we know the oxidation states of O and N we can find the oxidation of Pb as , $x+2\left( 5 \right)+6\left( -2 \right)=0$ $x+2\left( 5 \right)+6\left( -2 \right)=0$
$x=2$, hence we got the answer. Pb can have +4 and +2 oxidation states but using this method we can find which oxidation state it combines with other atoms.