Question

what element is shown in the electron configuration below? 1s²2s²2p⁶3s²3p⁶4s²3d¹⁰4p⁶5s¹
o cesium
o rubidium
o strontium
o potassium

Explanation:

Step1: Calculate total electrons

Add up the exponents. $2 + 2+6 + 2+6+2 + 10+6+1=37$.

Step2: Identify element by atomic number

The atomic number is equal to the number of electrons in a neutral atom. The element with atomic number 37 is Rubidium.

Answer:

Rubidium