Anomalous properties of nitrogen - class-XII

N-N sigma bond (single bod)  is weaker than P-P sigma bond (single bond) due to the small bond length between the nitrogen atoms. The lone pair of electrons of both the atoms nitrogen  repel each other making the single bon between $N-N$  weaker than P-P sigma bond.In $P-P$  bond the repulsion between sigma bond and lone pair electrons is less as the size of phosphorous is large.

So statement A is wrong.

Hence option A is correct.

$P$ has empty $d$ orbital, that can be filled by an additional electron. And since its size is bigger it is least likely to handle triple bond, as size increases $\pi$ overlapping decreases. 

In sulfur trioxide (SO3), sulfur utilizes its empty d-orbitals to form pi bonds with the p-orbitals of oxygen atoms, resulting in p_pi-d_pi bonding. The other species listed either lack vacant d-orbitals of appropriate energy or do not form d-pi bonds in the same manner.

$N(SiH _3) _3$ is planar due to back bonding and $N(CH _3) _3$ more basic due to presence of lone pair electrons.

The bond energy of $N _2$ per mole is 225 kcal or 945 KJ/ mol

Ammonium dichromate on heating gives nitrogen, chromic oxide and water.
$\displaystyle { \left( { NH } _{ 4 } \right)  } _{ 2 }{ Cr } _{ 2 }{ O } _{ 7 }\underrightarrow { \Delta  } { N } _{ 2 }+{ Cr } _{ 2 }{ O } _{ 3 }+4{ H } _{ 2 }O$

The atomic number of nitrogen is 7.

Phosphorus is unable to form $p\pi-p\pi$ bonds due to larger atomic size. So, $P$ atom is linked with $P$ atoms using $3$ sigma bonds. On the other hand, Nitrogen forms $1-\sigma$ and $2-\pi$ bonds i.e.,  triple bonds and exists as diatomic molecules.

Nitrogen is liberated by the thermal decomposition of ammonium nitrite $(NH _4NO _2), NaN _3, (NH _4) _2Cr _2O _7$

nitrogen show oxidation state of -1\3 in hydrazoic acid 

Bond order for $N _2=3$

In trisilylamine N(SiH3)3, the nitrogen uses its lone pair to form pπ-dπ back bonds with empty d-orbitals of silicon. This is similar to back bonding in trimethylamine N-oxide. Option B is incorrect (no count given), Option C's hybrid claim is questionable, and Option D is wrong - silanols are more acidic than alcohols.

$N _{2}O _{5}$ in solid form exists as $NO _{3}^{-}$ and $NO _{2}^{-}$.

N forms such bonds as,
$N \equiv N;  -C \equiv N;  N\equiv O$

As vacant d-orbital are present in P and other elements of this group except N so they can expand their valency and forms stable pentavalence product but due to absence of d-orbitals nitrogen cannot form pentavalence products.

Nitrogen differs from other members of the group due to its smaller size, high electronegativity, high IE and non availability of d-orbitals. Nitrogen has unique ability to form $p\pi - p\pi $ multiple bonds with itself and with other elements having small size and high electronegativity.

In atoms of nitrogen molecule, nonpolar covalent bonds are present.
Nonpolar covalent bond is formed between two atoms having zero electronegativty difference.

Nitrogen molecule ($\displaystyle N _2$) is relatively more stable. Each $N$ atom has completed its octet and there is a triple bond between two $N$ atoms. In other molecules, there is either a single bond or a double bond between two atoms. A triple bond is stronger than a single or double bond.

$NH _4Cl+NaNO _2 \rightarrow N _2+NaCl+2H _2O$

The weak catenation property of nitrogen is due to lone pair repulsion of lone pair of electrons on nitrogen molecule.

The single $N-N$ bonds formed is weaker than single $P-P$ bonds because of lone pair repulsion on small nitrogen molecule.

$2NaN _3 \rightarrow 2Na +3N _2$ 

Nitrogen forms oxides in which nitrogen exhibits each of its positive oxidation numbers from +1 to +5. $N _2O$ has nitrogen in the +1 oxidation state, $NO$ has +2 and $NO _2$ has +4 oxidation state.

Only highly electrnegative elements can make nitrogen exhibit +5 oxidation state

$N$ has 5 valence electron, therefore molecules with $N _2O _x$ will have even number of electrons and will be non radical. For monomeric oxide, $N _2O$ is the only monomeric nitrogen oxide.

The nitrogen which exists free in nature which is made up of 78% of Earth atmosphere and naturally found in mineral deposits, soil and organic compounds.

$N-N$ single bond is weaker than $P-P$ bond due to smaller size of $N$ as compared to $P$. Smaller size of $N$ leads to smaller $N-N$ bond length. As a result, the lone pair of electrons on the both the $N$ atoms repel each other leading to unstability or weakening of $N-N$ bond. Because of larger size of $P$ atom, $P-P$ bond length is more and lone pair-lone pair repulsion between $P$ atoms is less which makes the $P-P$ bond stronger than $N-N$ bond.

Alkali metal when reacts with liquid ammonia reaction is given by,

Hydrazine (N2H4) reacts with hydrogen peroxide as a rocket propellant, producing energy and gases for thrust. The equation suggests an oxidation reaction where H2O2 acts as an oxidizer. H2O2 is not used as bleaching agent, antiseptic, or for restoring paintings in this context.

Both $CN^-$ and $N _2$ have the same number of electrons but $N _2$ is more stable and chemically inert. This is because $N _2$ has a very high bond energy because of the presence of a triple bond between the two $N$ atoms. Both atoms in $N _2$ have achieved a stable electronic configuration, making the $N _2$ atom highly stable. $CN^-$ also has a triple bond but the bond is polar, unlike $N _2$ in which there is no bond polarity. Therefore, $CN^-$ is less stable and not chemically inert.

The only stable halide of nitrogen $NF _3$ is not explosive. The other halides are explosive due to their unstable nature because of much larger sizes.

Hydride of nitrogen is basic due to lone pair of electrons on nitrogen atom. The common hydrides of nitrogen are hydrazine,diazene,ammonia out of which only ammonia is thermally stable.

Nitrogen has very small size as compared to the halogens, which have much larger sizes. Due to this, they can not remain bonded to the nitrogen atom and hence are highly unstable. $NF _3 $ is stable because fluorine and nitrogen have comparable sizes and electronegativity.

In the commercial preparation of dinitrogen by the liquefaction and fractional distillation,dinitrogen is obtained in gaseous form.

$N _2+O _2 \rightarrow NO$

Nitrogen can combine with both metals and non metals.

$[Fe(H _2O) _5(NO)]SO _4$ is the complex formed in brown ring test done to identify the presence of nirates.