Question
Why $PCl _5$ exist and $NCl _5$ do not?
✓
Answer
Nitrogen is an element of second period which have 5 electrons in valence shell but have only 3 unpaired electrons.
Nitrogen do not have vacant d-orbital. Hence in this electron do not get in excited state. Hence in this octet do not get expanded, hence its valency can't be 5. Due to this $NCl _5$ do not exist. While P of $PCl _5$ is the element of third period. Hence in this electron get in excited state and valency get 5 and $PCl _5$ get easily formed.
Nitrogen do not have vacant d-orbital. Hence in this electron do not get in excited state. Hence in this octet do not get expanded, hence its valency can't be 5. Due to this $NCl _5$ do not exist. While P of $PCl _5$ is the element of third period. Hence in this electron get in excited state and valency get 5 and $PCl _5$ get easily formed.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Give an example of disproportionation reaction.
An alkane C8H18 is obtained as the only product on subjecting a primary alkyl halide to Wurtz reaction. On monobromination this alkane yields a single isomer of a tertiary bromide. Write the structure of alkane and the tertiary bromide.
List the following alkenes in decreasing order of reactivity towards electrophilic addition.
- ClCH2CH=CH2
- (CH3)2C=CH2
- CH3CH=CH2
- CH2=CHCl
Explain your order.
What will be value of (pKa + pKb) at 25°C?
Assuming complete dissociation, calculate the pH of the following solutions:
0.002 M HBr
At 523 K temperature following reaction $PCl _5(g) \rightleftharpoons PCl _3(g)+ Cl _2(g)$ value of $K _{ p }$ is 2.35 atm then calculate value of $K_c \cdot\left(R=0.082 L\right.$ atm degree ${ }^{-1}$ $mol ^{-1}$.
→Use the following thermodynamic data to calculate the enthalpy change for the formation of solid lithium fluoride, LiF(s) from Li(s) and F2(g):
→$\text{Li(s)}+\frac{1}{2}\text{F}_2(\text{g})\overrightarrow{\ \ \ \ \ \ \ }\ \text{LiF(s)}$
$\text{Li(s)}\overrightarrow{\ \ \ \ \ \ \ \ }\ \text{Li(g)};$ $\Delta_\text{s}\text{H}^\circ=155\text{kJ mol}^{-1}$
$\frac{1}{2}\text{F}_2(\text{g})\overrightarrow{\ \ \ \ \ \ \ }\ \text{F(g)};$ $\Delta\text{H}^\circ=75\text{kJ}$
$\text{Li(g)}\overrightarrow{\ \ \ \ \ } \ \text{Li}^{+}(\text{g})+\text{e}^-;$ $\Delta\text{H}^\circ=520\text{kJ mol}^{-1}$
$\text{F(g)}+\text{e}^-\overrightarrow{\ \ \ \ \ \ } \ \text{F}^-(\text{g});$ $\Delta\text{H}^\circ=-333\text{kJ mol}^{-1}$
$\text{Li}^+\text{(g)}+\text{F}^-(\text{g})\overrightarrow{\ \ \ \ \ \ \ \ }\ \text{LiF(s)};$ $\Delta\text{H}^\circ=-1012\text{kJ mol}^{-1}$
How will you convert:
- Propyne into propanone?
- Secondary butyl bromide into but-2-ene?
Write structural formulae for compounds named as:
- 1-bromoheptane.
- 5-bromoheptanoic acid.
Calculate the standard free energy change $\Delta\text{G}^\circ$ for the reaction: 2HgO(s) → 2Hg(l) + O2(g)
→$\Delta\text{H}^\circ=91\text{kJ mol}^{-1}$ at 298K, S°(HgO) = 72.0J K-1 mol-1, S°(Hg) = 77.4J K-1 mol-1 and $\text{S}^\circ_{(\text{O}_2)}=205\text{J K}^{-1}\text{mol}^{-1}$ at 298K.