Question

chem 1110 lab - naming and oxidation numbers
bonus activity - worth up to10 points
give the correct iupac name for the following compounds. give the oxidation number for the element listed.

1. nibr₂
2. h₂so₃(aq)
3. pb(clo₂)₄
4. cl₂o₅
5. uf₆
6. hno₃(aq)
7. (nh₄)₂s
8. agch₃coo
9. hi(aq)
10. cr(cn)₆
11. fe(io)₂
12. po₄³⁻
13. k₂cro₄
14. zn(oh)₂
15. hbro(aq)
16. in(clo₃)₃
17. na₂o
18. na₂o₂
19. cdco₃
20. (nh₄)₂so₄
21. moo₃
22. h₃po₃(aq)
23. ccl₄
24. so₃
25. io₂⁻

Explanation:

Step1: Recall oxidation - number rules

Oxidation number of hydrogen is +1 (except in metal hydrides), oxygen is -2 (except in peroxides and OF₂), and in a neutral compound, the sum of oxidation numbers is 0. In a poly - atomic ion, the sum of oxidation numbers equals the charge of the ion.

Step2: Analyze \(H_2SO_3\)

Let the oxidation number of \(S\) be \(x\). Since \(H = + 1\) and \(O=-2\), for \(H_2SO_3\): \(2\times( + 1)+x + 3\times(-2)=0\). Solving for \(x\): \(2 + x-6 = 0\), so \(x = + 4\).

Step3: Analyze \(Pb(ClO_2)_4\)

Let the oxidation number of \(Cl\) be \(y\). Oxygen has an oxidation number of -2. In the \(ClO_2^-\) ion, if we consider the ion's charge of - 1, then \(y+2\times(-2)=-1\), so \(y = + 3\).

Step4: Analyze \(Cl_2O_5\)

Let the oxidation number of \(Cl\) be \(z\). Since \(O=-2\), for \(Cl_2O_5\): \(2z+5\times(-2)=0\), so \(2z = 10\) and \(z = + 5\). The oxidation number of \(O\) is -2.

Step5: Analyze \(UF_6\)

Let the oxidation number of \(U\) be \(a\). Since \(F=-1\), for \(UF_6\): \(a + 6\times(-1)=0\), so \(a=+6\).

Step6: Analyze \(HNO_3\)

Let the oxidation number of \(N\) be \(b\). Since \(H = + 1\) and \(O=-2\), for \(HNO_3\): \(1 + b+3\times(-2)=0\), so \(b = + 5\).

Step7: Analyze \((NH_4)_2S\)

In \(NH_4^+\), let the oxidation number of \(N\) be \(c\). Since \(H = + 1\), then \(c + 4\times(+1)=+1\), so \(c=-3\). In \((NH_4)_2S\), the oxidation number of \(S\) is -2.

Step8: Analyze \(AgCH_3COO\)

The oxidation number of \(Ag\) is +1.

Step9: Analyze \(HI\)

The oxidation number of \(I\) is -1.

Step10: Analyze \(Cr(CN)_6\)

In \(CN^-\), the oxidation number of \(C\) is +2 and \(N\) is -3. Let the oxidation number of \(Cr\) be \(d\). Since \(CN^-\) has a charge of -1, for \(Cr(CN)_6\): \(d+6\times(-1)=0\), so \(d = + 6\).

Step11: Analyze \(Fe(IO)_2\)

In \(IO^-\), let the oxidation number of \(I\) be \(e\). Since \(O=-2\), then \(e+( - 2)=-1\), so \(e = + 1\).

Step12: Analyze \(PO_4^{3 -}\)

Let the oxidation number of \(P\) be \(f\). Since \(O=-2\), for \(PO_4^{3 -}\): \(f+4\times(-2)=-3\), so \(f = + 5\).

Step13: Analyze \(K_2CrO_4\)

Let the oxidation number of \(Cr\) be \(g\). Since \(K = + 1\) and \(O=-2\), for \(K_2CrO_4\): \(2\times(+1)+g + 4\times(-2)=0\), so \(g = + 6\).

Step14: Analyze \(Zn(OH)_2\)

The oxidation number of \(Zn\) is +2.

Step15: Analyze \(HBrO\)

Let the oxidation number of \(Br\) be \(h\). Since \(H = + 1\) and \(O=-2\), for \(HBrO\): \(1+h+( - 2)=0\), so \(h = + 1\).

Step16: Analyze \(In(ClO_3)_3\)

Let the oxidation number of \(In\) be \(i\). In \(ClO_3^-\), let the oxidation number of \(Cl\) be \(j\). Since \(O=-2\), for \(ClO_3^-\): \(j + 3\times(-2)=-1\), so \(j = + 5\). For \(In(ClO_3)_3\), \(i+3\times(-1)=0\), so \(i = + 3\).

Step17: Analyze \(Na_2O\)

The oxidation number of \(O\) is -2 and \(Na\) is +1.

Step18: Analyze \(Na_2O_2\)

In \(Na_2O_2\), since \(Na = + 1\), let the oxidation number of \(O\) be \(k\). For \(Na_2O_2\): \(2\times(+1)+2k = 0\), so \(k=-1\).

Step19: Analyze \(CdCO_3\)

The oxidation number of \(Cd\) is +2.

Step20: Analyze \((NH_4)_2SO_4\)

In \(NH_4^+\), \(N=-3\) and \(H = + 1\). In \(SO_4^{2 -}\), let the oxidation number of \(S\) be \(m\). Since \(O=-2\), for \(SO_4^{2 -}\): \(m+4\times(-2)=-2\), so \(m = + 6\). The oxidation number of \(H\) in \(NH_4^+\) is +1.

Step21: Analyze \(MoO_3\)

Let the oxidation number of \(Mo\) be \(n\). Since \(O=-2\), for \(MoO_3\): \(n+3\times(-2)=0\), so \(n = + 6\).

Step22: Analyze \(H_3PO_3\)

Let the oxidation number of \(P\) be \(p\). Since \(H = + 1\) and \(O=-2\), for \(H_3PO_3\): \(3\times(+1)+p+3\times(-2)=0\), so \(p = + 3\). The oxidation number of \(H\) is +1.

Step23: Analyze \(CCl_4\)

Let the oxidation number of \(C\) be \(q\). Since \(Cl=-1\), for \(CCl_4\): \(q+4\times(-1)=0\), so \(q = + 4\). The oxidation number of \(Cl\) is -1.

Step24: Analyze \(SO_3\)

Let the oxidation number of \(S\) be \(r\). Since \(O=-2\), for \(SO_3\): \(r+3\times(-2)=0\), so \(r = + 6\).

Step25: Analyze \(IO_2^-\)

Let the oxidation number of \(I\) be \(s\). Since \(O=-2\), for \(IO_2^-\): \(s+2\times(-2)=-1\), so \(s = + 3\).

Answer:

1. Nickel(II) bromide; \(Ni = + 2\)
2. Sulfurous acid; \(S = + 4\)
3. Lead(IV) chlorite; \(Cl = + 3\)
4. Dichlorine pentoxide; \(Cl = + 5\), \(O=-2\)
5. Uranium(VI) fluoride; \(U = + 6\)
6. Nitric acid; \(N = + 5\)
7. Ammonium sulfide; \(N=-3\), \(S=-2\)
8. Silver acetate; \(Ag = + 1\)
9. Hydroiodic acid; \(I=-1\)
10. Chromium(VI) cyanide; \(Cr = + 6\)
11. Iron(II) hypoiodite; \(I = + 1\)
12. Phosphate ion; \(P = + 5\)
13. Potassium chromate; \(Cr = + 6\)
14. Zinc hydroxide; \(Zn = + 2\)
15. Hypobromous acid; \(Br = + 1\)
16. Indium(III) chlorate; \(Cl = + 5\)
17. Sodium oxide; \(O=-2\)
18. Sodium peroxide; \(O=-1\)
19. Cadmium carbonate; \(Cd = + 2\)
20. Ammonium sulfate; \(N=-3\), \(S = + 6\), \(H = + 1\)
21. Molybdenum(VI) oxide; \(Mo = + 6\)
22. Phosphorous acid; \(P = + 3\), \(H = + 1\)
23. Carbon tetrachloride; \(C = + 4\), \(Cl=-1\)
24. Sulfur trioxide; \(S = + 6\)
25. Iodite ion; \(I = + 3\)