Transition and InnerTransition Elements - Result Question 77
####3. The correct order of the first ionisation enthalpies is
(2019 Main, 10 April II)
(a) $Mn<Ti<Zn<Ni$
(b) $Ti<Mn<Zn<Ni$
(c) $Zn<Ni<Mn<Ti$
(d) $Ti<Mn<Ni<Zn$
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Solution:
- The $3 d$-transition series is
$$ \begin{array}{lccccccccccc} & Sc & Ti & T & Cr & Mn & Fe & Co & Ni & Cu & Zn \ & \text { Atomic number } & 21 & 22 & 23 & 24 & 25 & 26 & 27 & 28 & 29 & 30 \ & & \Downarrow & & \Downarrow & & & \Downarrow & & \Downarrow \ & & \Downarrow & & & & & \ \text { Outermost } & & 3 d^{2} 4 s^{2} & & 3 d^{5} 4 s^{2} & & 3 d^{8} 4 s^{2} & 3 d^{10} 4 s^{2} \ \text { Electronic } & & & & & & & \ \text { Configuration } & & & & & & & \end{array} $$
In 1st ionisation, one electron will be removed from $4 s^{2}$ subshell/orbital.
With increase in atomic number $(Z)$, i.e. with increase in number of protons in the nucleus, effective nuclear charge $\left(Z^{*}\right)$ also increases from $Sc$ to $Zn$.
$$ IE \propto Z^{*} $$
So, IE order of the given elements will be,
$$ Ti<Mn<Ni<Zn $$
