COORDINATION COMPOUND PRACTICE WORKSHEET

 

Class 12 Chemistry – Coordination Compounds

A. Effective Atomic Number (EAN) – 5 Questions

1. Calculate the Effective Atomic Number (EAN) of Co in $[Co(NH_3)_6]Cl_3$.

2. Calculate the EAN of Fe in $K_4[Fe(CN)_6]$.

3. Calculate the EAN of Fe in $K_3[Fe(CN)_6]$.

4. Calculate the EAN of Cu in $[Cu(NH_3)_4]SO_4$.

5. Calculate the EAN of Cr in $K_3[Cr(C_2O_4)_3]$.

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B. IUPAC Naming – 10 Questions

Write the IUPAC name of the following coordination compounds:

1. $[Co(NH_3)_6]Cl_3$

2. $K_4[Fe(CN)_6]$

3. $K_3[Fe(CN)_6]$

4. $[Cu(NH_3)_4]SO_4$

5. $[Ag(NH_3)_2]Cl$

6. $K_2[PtCl_6]$

7. $[Cr(NH_3)_4Cl_2]Cl$

8. $[Co(NH_3)_5Cl]Cl_2$

9. $Na_2[Zn(OH)_4]$

10. $[Ni(CO)_4]$

C. Challenge Questions – IUPAC Naming

11. $[Co(en)_3]Cl_3$

12. $K_3[Cr(C_2O_4)_3]$

13. $[Pt(NH_3)_2Cl_2]$

14. $[Co(NH_3)_4Cl_2]Cl$

15. $N{a}_3[Al{F}_6]$

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ANSWERS

Class 12 Chemistry – Coordination Compounds

Questions with Solutions

A. Effective Atomic Number (EAN) – 5 Questions

Formula:
EAN = Atomic number of metal − Oxidation state + Electrons donated by ligands

1. Find the EAN of Co in $[Co(NH_3)_6]Cl_3$.

Solution:
Atomic number of Co = 27
Oxidation state of Co = +3
6 $NH_3$ ligands donate $6 \times 2 = 12$ electrons.

$$EAN=27-3+12=\boxed{36}$$

Answer: 36

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2. Find the EAN of Fe in $K_4[Fe(CN)_6]$.

Solution:
Oxidation state of Fe:

$$x+6(-1)=-4$$ $$x=+2$$

Atomic number of Fe = 26
Six $CN^-$ ligands donate 12 electrons.

$$EAN=26-2+12=\boxed{36}$$

Answer: 36

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3. Find the EAN of Fe in $K_3[Fe(CN)_6]$.

Solution:

$$x+6(-1)=-3$$ $$x=+3$$ $$EAN=26-3+(6\times2)$$ $$=\boxed{35}$$

Answer: 35

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4. Find the EAN of Cu in $[Cu(NH_3)_4]SO_4$.

Solution:
$SO_4^{2-}$ is outside the coordination sphere, so the complex ion has a charge of $+2$.

Therefore, Cu oxidation state = +2.

Atomic number of Cu = 29

Four $NH_3$ ligands donate:

$$4\times2=8$$ $$EAN=29-2+8=\boxed{35}$$

Answer: 35

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5. Find the EAN of Cr in $K_3[Cr(C_2O_4)_3]$.

Solution:
Oxidation state of Cr:

$$x+3(-2)=-3$$ $$x=+3$$

Atomic number of Cr = 24.

Each $C_2O_4^{2-}$ is a bidentate ligand and donates 4 electrons.

Therefore, 3 oxalate ligands donate:

$$3\times4=12$$ $$EAN=24-3+12=\boxed{33}$$

Answer: 33

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B. IUPAC Naming – 10 Questions with Solutions

1. $[Co(NH_3)_6]Cl_3$

Solution:
Ligand = $NH_3$ → ammine
Number = 6 → hexaammine
Co oxidation state = +3

IUPAC name:
Hexaamminecobalt(III) chloride

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2. $K_4[Fe(CN)_6]$

Fe oxidation state = +2.

$CN^-$ → cyanido
6 ligands → hexacyanido
Since complex ion is anionic, Fe → ferrate

IUPAC name:
Potassium hexacyanidoferrate(II)

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3. $K_3[Fe(CN)_6]$

Fe oxidation state = +3.

IUPAC name:
Potassium hexacyanidoferrate(III)

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4. $[Cu(NH_3)_4]SO_4$

Cu oxidation state = +2.

IUPAC name:
Tetraamminecopper(II) sulfate

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5. $[Ag(NH_3)_2]Cl$

Ag oxidation state = +1.

2 $NH_3$ → diammine

IUPAC name:
Diamminesilver(I) chloride

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6. $K_2[PtCl_6]$

Oxidation state of Pt:

$$x+6(-1)=-2$$ $$x=+4$$

6 $Cl^-$ → hexachlorido
Anionic complex → platinate

IUPAC name:
Potassium hexachloridoplatinate(IV)

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7. $[Cr(NH_3)_4Cl_2]Cl$

The complex ion has charge $+1$.

$$x+4(0)+2(-1)=+1$$ $$x=+3$$

4 $NH_3$ → tetraammine
2 $Cl^-$ → dichlorido

IUPAC name:
Tetraamminedichloridochromium(III) chloride

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8. $[Co(NH_3)_5Cl]Cl_2$

Complex ion charge = +2.

$$x+5(0)-1=+2$$ $$x=+3$$

IUPAC name:
Pentaamminechloridocobalt(III) chloride

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9. $Na_2[Zn(OH)_4]$

Oxidation state of Zn:

$$x+4(-1)=-2$$ $$x=+2$$

4 $OH^-$ → tetrahydroxido
Anionic complex → zincate

IUPAC name:
Sodium tetrahydroxidozincate(II)

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10. $[Ni(CO)_4]$

CO is a neutral ligand.

Oxidation state of Ni = 0.

4 CO → tetracarbonyl

IUPAC name:
Tetracarbonylnickel(0)

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C. Challenge Questions – Solutions

11. $[Co(en)_3]Cl_3$

$en$ = ethane-1,2-diamine, a neutral bidentate ligand.

Co = +3.

IUPAC name:
Tris(ethane-1,2-diamine)cobalt(III) chloride

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12. $K_3[Cr(C_2O_4)_3]$

Cr = +3 and oxalate is bidentate.

IUPAC name:
Potassium tris(oxalato)chromate(III)

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13. $[Pt(NH_3)_2Cl_2]$

Pt oxidation state:

$$x-2=0$$ $$x=+2$$

IUPAC name:
Diamminedichloridoplatinum(II)

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14. $[Co(NH_3)_4Cl_2]Cl$

Complex charge = +1.

$$x-2=+1$$ $$x=+3$$

IUPAC name:
Tetraamminedichloridocobalt(III) chloride

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15. $Na_3[AlF_6]$

Oxidation state of Al:

$$x+6(-1)=-3$$ $$x=+3$$

IUPAC name:
Sodium hexafluoridoaluminate(III)