Q.Match Column I with Column II for the oxidation states of the central atoms.
Column I

The key concept here is oxidation state determination in unusual compounds. Many students see CrO₅ and assume oxygen is always –2, leading to a wrong oxidation state for chromium. But CrO₅ contains peroxide linkages (O–O bonds), where oxygen has oxidation state –1, not –2. Recognizing this is the entire crux of the problem.

Let’s work through it step by step.

1. Find the oxidation state of Cr in the reactant Cr₂O₇²⁻ In dichromate ion, each oxygen is –2 (no peroxide bonds). Let Cr have oxidation state $x$.

So Cr is +6 in the reactant.

2. Now examine the product CrO₅ CrO₅ is known as perchromic acid (or chromium pentoxide). Its structure is not five separate O²⁻ ions; instead, it contains one oxo (Cr=O) and two peroxide (O–O) groups.
   • In a peroxide bond, each oxygen has oxidation state –1.
   • In the oxo group, oxygen is –2. So in CrO₅:
   • 1 oxygen as oxo: –2
   • 4 oxygens as two peroxides: each peroxide contributes –2 total (since each O is –1, two O’s give –2), so two peroxides give –4. Total oxygen contribution = –2 + (–4) = –6. Let Cr oxidation state be $y$. The molecule is neutral:
   $$$$

The key concept here is that alkali metals (Group 1) have a fixed oxidation number of +1 in all their compounds. This is because they readily lose their single valence electron to achieve a noble gas configuration. Oxygen, on the other hand, can have variable oxidation numbers depending on the type of compound: -2 in normal oxides, -1 in peroxides, and -1/2 in superoxides. The trick is not to let oxygen’s variability confuse you—potassium’s oxidation state is constant.

Let’s work through each compound step by step:

1. In $\mathrm{K}_2 \mathrm{O}$ (potassium oxide):

   • Potassium (K) is in Group 1, so its oxidation number is always $+1$.
   • Let the oxidation number of oxygen be $x$. The compound is neutral, so: $2(+1) + x = 0 \implies x = -2$.
   • This is the normal oxide ion $\mathrm{O}^{2-}$. Potassium’s oxidation number is $+1$.

2. In $\mathrm{K}_2 \mathrm{O}_2$ (potassium peroxide):

   • Again, potassium is $+1$.
   • Let oxygen’s oxidation number be $y$. The sum of oxidation numbers is zero: $2(+1) + 2y = 0 \implies 2y = -2 \implies y = -1$.
   • This is the peroxide ion $\mathrm{O}_2^{2-}$, where each oxygen has an oxidation number of $-1$. Potassium remains $+1$.

3. In $\mathrm{KO}_2$ (potassium superoxide):

   • Potassium is still $+1$. …

The key idea is to assign oxidation numbers systematically using the rule that oxygen is almost always −2 (except in peroxides, which don’t appear here), and the overall charge of the ion equals the sum of oxidation numbers. Once we compute each central atom’s oxidation state, we simply sort them from highest to lowest.

Why this works: Oxidation number is a bookkeeping tool that tracks electron distribution in compounds and ions. For polyatomic ions, the sum of oxidation numbers equals the ion’s charge. Oxygen’s fixed −2 makes these calculations straightforward.

1. Find oxidation number of Cr in $\mathrm{Cr}_2\mathrm{O}_7^{2-}$

   Let Cr’s oxidation number be $x$. There are 2 Cr atoms and 7 O atoms (each O = −2). The ion charge is −2.

   Equation: $2x + 7(-2) = -2$

   $2x - 14 = -2 \implies 2x = 12 \implies x = +6$.

   So Cr here is +6.

2. Find oxidation number of Cr in $\mathrm{CrO}_2^{-}$

   Let Cr’s oxidation number be $y$. One Cr, two O atoms (each −2), charge −1.

   $y + 2(-2) = -1 \implies y - 4 = -1 \implies y = +3$.

   So Cr here is +3.

3. Find oxidation number of Mn in $\mathrm{MnO}_4^{-}$

   Let Mn’s oxidation number be $z$. One Mn, four O atoms (each −2), charge −1.

   $z + 4(-2) = -1 \implies z - 8 = -1 \implies z = +7$.

   So Mn is +7.

4. Find oxidation number of Br in $\mathrm{BrO}_3^{-}$

   Let Br’s oxidation number be $w$. One Br, three O atoms (each −2), charge −1.

   $w + 3(-2) = -1 \implies w - 6 = -1 \implies w = +5$.

   So Br is +5. …

We are given the oxidation numbers:

A = +3, B = +6, C = –2.

For a neutral compound, the sum of (oxidation number × number of atoms) over all elements must equal zero.

Let’s check each option step by step.

1. Option (A): $\mathrm{A}_3\left(\mathrm{BC}_3\right)_4$

   • This means 3 atoms of A, and 4 formula units of $\mathrm{BC}_3$.
   • In one $\mathrm{BC}_3$: B = +6, each C = –2, so total for $\mathrm{BC}_3$ = $+6 + 3(-2) = 0$.
   • So the polyatomic ion $\mathrm{BC}_3$ is neutral.
   • Then the compound is just 3 A atoms: total charge = $3 \times (+3) = +9$.
   • Not zero → not neutral.

2. Option (B): $\mathrm{A}_2\left(\mathrm{BC}_4\right)_3$

   • 2 atoms of A, and 3 formula units of $\mathrm{BC}_4$.
   • In one $\mathrm{BC}_4$: B = +6, each C = –2, so total = $+6 + 4(-2) = +6 - 8 = -2$.
   • So $\mathrm{BC}_4$ has a charge of –2.
   • Then total charge = $2(+3) + 3(-2) = +6 - 6 = 0$.
   • This works.

3. Option (C): $\mathrm{A}_3(\mathrm{BC})_2$

   • 3 atoms of A, and 2 formula units of $\mathrm{BC}$.
   • In one $\mathrm{BC}$: B = +6, C = –2, total = $+6 - 2 = +4$.
   • Total charge = $3(+3) + 2(+4) = +9 + 8 = +17$.
   • Not zero. …

Concept: Oxidation state in a metal carbonyl. Carbon monoxide is a NEUTRAL ligand (charge 0).

For [Ni(CO)4] (tetracarbonylnickel(0)), the complex itself is neutral:

x + 4(0) = 0 => x = 0 …

Concept: Oxidation number with peroxo linkages.

From the figure, CrO5 (chromium peroxide, the blue species of the H2O2/dichromate test) has a butterfly structure:

• ONE terminal oxo oxygen, Cr=O -> oxidation state of that O = -2
• TWO peroxo (O-O) groups bonded side-on -> each O in a peroxide linkage has oxidation state -1, i.e. 4 oxygens at -1 …

Counting bonds (C–H = $-1$ to C, C–O = $+1$, C=O = $+2$, C–C = $0$):

• $C_2H_6$: each C has 3 C–H + 1 C–C $= -3$.
• $CH_3OH$: 3 C–H + 1 C–O $= -3 + 1 = -2$.
• $CH_2O$: 2 C–H + 1 C=O $= -2 + 2 = 0$. …

Oxidation state of Mn in each product (from +7):

• Potassium manganate K$_2$MnO$_4$: Mn = +6 $\Rightarrow$ 1 e⁻
• Manganese dioxide MnO$_2$: Mn = +4 $\Rightarrow$ 3 e⁻ …

Step 1 — Write the formulae of the four copper ores.

Step 2 — Assign oxidation states.

Cuprite, $Cu_2O$: oxygen is $-2$, and the compound is neutral, so

Copper is +1 (cuprous).

Malachite, $CuCO_3\cdot Cu(OH)_2$: in $CuCO_3$, the carbonate ion is $CO_3^{2-}$, so $Cu$ is $+2$; in $Cu(OH)_2$, two $OH^-$ give $Cu = +2$. …

The key idea is that hybridisation is a local property of an atom in a molecule — it depends only on how many regions of electron density (sigma bonds + lone pairs) surround that atom. Each region corresponds to one hybrid orbital. So to find the hybridisation of carbon in each species, we simply count the number of atoms directly bonded to carbon plus any lone pairs on carbon, and then match that number to the familiar hybridisation scheme: 2 regions → $sp$, 3 regions → $sp^2$, 4 regions → $sp^3$.

Let’s apply this to each molecule in turn.

1. $\mathrm{CO_2}$ (carbon dioxide)

   Carbon is the central atom. It forms two double bonds — one with each oxygen. Each double bond counts as one sigma bond and one pi bond. So carbon has exactly two sigma bonds and no lone pairs. That gives 2 regions of electron density.

   Two regions → $sp$ hybridisation.

   Tip

   A double bond does NOT count as two regions — it’s still just one direction in space. Only sigma bonds and lone pairs create separate hybrid orbitals.

2. $\mathrm{CH_4}$ (methane)

   Carbon is bonded to four hydrogen atoms by four single bonds. Each single bond is one sigma bond. So carbon has four sigma bonds and no lone pairs. That’s 4 regions.

   Four regions → $sp^3$ hybridisation.

3. $\mathrm{CO_3^{2-}}$ (carbonate ion) …