Join WhatsApp Icon JEE WhatsApp Group
Question 43

Nd$$^{2+}$$ = ______

Neodymium (Nd) has atomic number 60.

Electronic configuration of Nd:

$$ \text{Nd} = [\text{Xe}] \; 4f^4 \; 6s^2 $$

For Nd$$^{2+}$$, two electrons are removed. In lanthanides, the $$6s$$ electrons are removed first (as they are in the outermost shell):

$$ \text{Nd}^{2+} = [\text{Xe}] \; 4f^4 $$

Get AI Help

Create a FREE account and get:

  • Free JEE Mains Previous Papers PDF
  • Take JEE Mains paper tests

50,000+ JEE Students Trusted Our Score Calculator

Predict your JEE Main percentile, rank & performance in seconds

Ask AI

Ask our AI anything

AI can make mistakes. Please verify important information.