Which of the following has more unpaired d electrons?

^{+}

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

^{2 +}

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

^{+}

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

^{+}

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

Open in App

Solution

The correct option is A Fe $^{2 +}$
No. of unpaired d electrons in the given ions is as follow:
$Z n^{+} - 1$
$F e^{2 +} - 4$
$N i^{+} - 3$
$C u^{+} - 0$
So, $F e^{2 +}$ has maximum number of unpaired electrons among the given elements.

Suggest Corrections

Similar questions

e^{-}

S c = 21, F e = 26, V = 23, N i = 28)

S c = 21, F e = 26, V = 23, N i = 28)

F e^{+ 3}

F e^{+ 2}

F e^{+ 3}

F e^{+ 2}

[F e (H_{2} O)_{6}]^{2 +}

[F e (C N)_{6}]^{4 -}

[N i (C O)_{4}]

[N i (H_{2} O)_{4}]^{2 +}

[N i (C N)_{4}]^{2 -}

Join BYJU'S Learning Program