Atomic size of nickel, cobalt and iron is in order of : $F e > C o > N i$ .

True

No worries! We‘ve got your back. Try BYJU‘S free classes today!

False

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is B False
As we move along a period, radius decreases but in $d - b l o c k$ they have almost same radius or can say they have approximately equal size.

Suggest Corrections

Similar questions

$[N i (C O)_{4}] > [C o (C O)_{4}]^{-} > [F e (C O)_{4}]^{2 -}$ is the correct order with respect to M-C $π$ bond. Where M=Ni, Fe, and Co

298 K

E_{N i^{2 +} (a q) / N i (s)}^{\circ} = - 0.24 V

E_{F e^{2 +} (a q) / F e (s)}^{\circ} = - 0.44 V

Join BYJU'S Learning Program