Zn\(^{2+}\) salts are white while Cu\(^{2+}\) salts are blue in colour. Why?
Step 1: Electronic Configurations: - Zn\(^{2+}\) = \(3d^{10}\) (fully filled, no d-d transitions)
- Cu\(^{2+}\) = \(3d^9\) (one unpaired electron, d-d transitions possible)
Step 2: Reason for Colour: - Zn\(^{2+}\) has no unpaired d-electrons → No d-d transitions → Appears white.
- Cu\(^{2+}\) has d-d transitions absorbing red-orange light, reflecting blue.
Consider the following reactions $ A + HCl + H_2SO_4 \rightarrow CrO_2Cl_2$ + Side Products Little amount $ CrO_2Cl_2(vapour) + NaOH \rightarrow B + NaCl + H_2O $ $ B + H^+ \rightarrow C + H_2O $ The number of terminal 'O' present in the compound 'C' is ______
An aqueous solution of Co(ClO4)2·6H2O is light pink in colour. Addition of conc. HCl results in an intense blue coloured solution due to the formation of a new species. The new species among the following is:
[Given: Atomic number of Co = 27]
Nature of compounds TeO₂ and TeH₂ is___________ and ______________respectively.
(b) Order of the differential equation: $ 5x^3 \frac{d^3y}{dx^3} - 3\left(\frac{dy}{dx}\right)^2 + \left(\frac{d^2y}{dx^2}\right)^4 + y = 0 $