1c

Question 1

LO 47 / 48

Answer 1

Question 2

LO 57

Answer 2

Ag+ ions would form precipitate with free Cl^- ions so there had to be 1 out of 3 ( ⅓) of the chlorines existing as free ions. The others are ligands on the metal complex ion.

Question 3

Answer 3

Question 4

LO 63 / 64 / 65

Answer 4

Question 5

LO 59 / 60 / 61

Answer 5

In an isolated Fe²⁺ ion

Question 6

Answer 6

in an octahedral complex with ligands splitting d orbitals

Question 7

LO 45

Answer 7

A: Same number of occupied levels

B: Fe ³⁺ has more unpaired electrons

C: Lower oxidation number of metals the better they are as reducing agents

D: Fe³⁺ has a stable half-full d shell

Question 8

LO 47 / 48

Answer 8

Question 9

LO 54 / 55

Answer 9

The copper is “bonding” at four points so has coordination number of 4.

The ligand is sharing (donating) 4 pairs of electrons points so is tetradentate

Question 10

LO 63 / 64 / 65

Answer 10

Ni has electron arrangement of [Ar] 4s² 3d⁸

The 4s electrons are lost first when the Ni²⁺ ion is formed so it will still have 3d⁸

C and D cannot be possible as there will be splitting of d orbitals due to ligands

Question 11

LO 57

Answer 11

7

The coordination number is the total number of bonds of ligand(s) to the central transition metal ion.

Question 12

LO 56

Answer 12

amminepentaaquanickel(II)

momoamminepentaaquanickel(II)

Question 13

LO 54 / 55

Answer 13

Forms only one bond (to the metal)
OR
donates/shares one (lone/nonbonding) pair of electrons

Question 14

LO 46

Answer 14

D

Question 15

LO 46

Answer 15

B

Question 16

LO 46

Answer 16

A: only has 3 unpaired

B: 1s 2s and 3s occupied but that gives only 6 electrons

C: 3s² 3p⁶ 3d⁷ = 16

D: Correct

Question 17

LO 47 / 48

Answer 17

A: +1

B: +3

C: +5

D: +7