Explanation
A neutral sodium atom is likely to achieve an octet in its outermost shell by losing its one valence electron. The cation produced in this way is Na$$^+$$, having electronic configuration [Ne] 3s$$^0$$ which means there is no unpaired electron.
Hence, Na is diamagnetic in its existing ion form.
A reduction in atomic/ionic size with increase in atomic number is a characteristic of lanthanoids. Hence, the ionic radius of Lu3+ will be smaller than the ionic radius of La3+ It will be less than 1.06 Å. Hence, the option C, 0.85 Å is the correct answer.
Actinoids are element 89 to 103 and fill their $$5 \mathrm{f}$$ shell. They are typical metals and radioactive also. $$E g-T h, U, P u$$.
$$U(z=92) \rightarrow 5 f^{3} 6 d^{1} 7 s^{2}$$
$$N{p}(z=93) \rightarrow 5 f^{4} 6 d^{1} 7 s^{2}$$
$$P u(z=94) \rightarrow 5 f^{6} 6 d^{0} 7 s^{2}$$
$$A m(z=95) \rightarrow 5 f^{7} 6 d^{\circ} 7 s^{2}$$
So, U and $$Np$$ have one electron in $$6 d$$ orbital.
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