To determine which compound is paramagnetic, we need to examine the oxidation state of manganese and the electron configuration of the manganese ion in each compound. - KMnO$_4$ (Potassium permanganate): In KMnO$_4$, the manganese ion exists in the \(+7\) oxidation state (Mn$^{7+}$). The electron configuration of Mn$^{7+}$ is \( 3d^0 4s^0 \), meaning that all d-orbitals are empty. Since there are no unpaired electrons, KMnO$_4$ is diamagnetic.
- K$_2$MnO$_4$ (Potassium manganate): In K$_2$MnO$_4$, the manganese ion exists in the \(+6\) oxidation state (Mn$^{6+}$). The electron configuration of Mn$^{6+}$ is \( 3d^1 4s^0 \), meaning there is one unpaired electron in the d-orbital. Since there is at least one unpaired electron, K$_2$MnO$_4$ is paramagnetic.
Step 1: Examine the oxidation state of manganese in each compound.
Step 2: Check the number of unpaired electrons in the electron configuration of the manganese ion.
Write IUPAC names of the following compounds and classify them into primary, secondary and tertiary amines.
(i) (CH3 )2CHNH2 (ii) CH3 (CH2 )2NH2 (iii) CH3NHCH(CH3 )2
(iv) (CH3 )3CNH2 (v) C6H5NHCH3 (vi) (CH3CH2 )2NCH3 (vii) m–BrC6H4NH2
Give one chemical test to distinguish between the following pairs of compounds.
(i) Methylamine and dimethylamine
(ii) Secondary and tertiary amines
(iii) Ethylamine and aniline
(iv) Aniline and benzylamine
(v) Aniline and N-methylaniline
Account for the following:
(i) pKb of aniline is more than that of methylamine.
(ii) Ethylamine is soluble in water whereas aniline is not.
(iii) Methylamine in water reacts with ferric chloride to precipitate hydrated ferric oxide.
(iv) Although amino group is o– and p– directing in aromatic electrophilic substitution reactions, aniline on nitration gives a substantial amount of m-nitroaniline.
(v) Aniline does not undergo Friedel-Crafts reaction.
(vi) Diazonium salts of aromatic amines are more stable than those of aliphatic amines. (vii) Gabriel phthalimide synthesis is preferred for synthesising primary amines.
Acidified \(KMnO_4\) oxidizes sulphite to:
Complete and balance the following chemical equations: (a) \[ 2MnO_4^-(aq) + 10I^-(aq) + 16H^+(aq) \rightarrow \] (b) \[ Cr_2O_7^{2-}(aq) + 6Fe^{2+}(aq) + 14H^+(aq) \rightarrow \]
Which element is a strong reducing agent in +2 oxidation state and why?
Convert Propanoic acid to Ethane