Question

Write the full nuclear symbol (elemental symbol with atomic number and mass number) for an isotope of Zirconium having 51 neutrons. I got (91/41)Zr

b. This element forms an oxide of having the formula ZrO2, called zirconia. What is the charge on Zirconium in this compound? I got +4

c. What is the electron configuration, valence shell orbital diagram and noble gas configuration for the Zirconium ion in ZrO2? I got 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4d6

Electron configuration: ________________________________________

Noble gas configuration: [ ] __________________________________ [Ar] Valence shells: 4s __ __ __ __ __ __ __ __ __ do not understand this

d. Is neutral zirconium paramagnetic? How many unpaired electrons?

e. Is the ion in part b paramagnetic? How many unpaired electrons? (2 points)

Answer 1

(a) Atomic number of Zr is 40. Hence there are 40 protons and 40 electrons in Zr.

Given that there are 51 neutrons. Hence mass number of Zr is

mass number, A = Number of protons + number of neutrons = 40 + 51 = 91

Hence neuclear symbol for Zr is (91/40) Zr (answer)

(b): Since ZrO2 is an oxide, the oxidation state of O is -2.

=> x + 2x*(-2) = 0

=> x = +4

(b) is correct

(c) Electronic configuration: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d² 5s²

While writing noble gas configuration, we write the nearst before noble gas in bracket and the rest of the electrons are filled as usual. Hence

Noble gas configuration: [Kr] 4d² 5s²

Valence shell orbital diagram

:

(d): Since there are 2 unpaired electrons in neutral zirconium atom, it is paramagnetic.

(e) in Zr⁴⁺ all the 4 valence electrons are lost during the formation of the cation. Hence there are no unpaired electron. Hence the ion is not paramagnetic.