- ✓$Fe$
- B$Ni$
- C$Co$
- D$Mn$
$\quad \quad \quad \quad _{25} \mathrm{Mn} \quad \quad \quad _{26} \mathrm{Fe} \quad \quad \quad _{27} \mathrm{Co} \quad \quad \quad \quad _{28}Ni$
$\mathrm{M}= \quad [\mathrm{Ar}] 3 \mathrm{d}^{3} 4 \mathrm{s}^{2} \quad[\mathrm{Ar}] 3 \mathrm{d}^{6} 4 \mathrm{s}^{2} \quad[\mathrm{Ar}] 3 \mathrm{d}^{7} 4 \mathrm{s}^{2} \quad[\mathrm{Ar}] 3 \mathrm{d}^{2} 4 \mathrm{s}^{2}$
$\mathrm{M}^{2+}=[\mathrm{Ar}] 3 \mathrm{d}^{3} 4 \mathrm{s}^{0} \quad[\mathrm{Ar}] 3 \mathrm{d}^{6} 4 \mathrm{s}^{0} \quad[\mathrm{Ar}] 3 \mathrm{d}^{7} 4 \mathrm{s}^{0} \quad[\mathrm{Ar}] 3 \mathrm{d}^{2} 4 \mathrm{s}^{6}$
So third ionisation energy is minimum for $Fe$
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Hyperconjugation occurs in
$C _2 H _4( g )+3 O _2( g ) \rightarrow 2 CO _2( g )+2 H _2 O ( l )$ the amount of heat produced as measured in bomb calorimeter is $1406\,kJ\,mol { }^{-1}$ at $300\,K$. The minimum value of $T \Delta S$ needed to reach equilibrium is $(-).......\,kJ$. (Nearest integer) Given : $R =8.3\,JK ^{-1} mol ^{-1}$
