The number of s-electrons present in an ion with 55 protons in its unipositive state is
To count s-electrons, include all electrons in the s-subshells up to the element’s configuration.
The element with 55 protons is cesium (Cs), and its atomic number is Z = 55. The electronic configuration of neutral Cs is:
\(Cs : [Xe] 6s ^1\)
For the unipositive ion Cs+, it loses the single 6s electron, and the configuration becomes:
\(Cs^+: [Xe]\)
Now count the total number of s-electrons in\( Cs^+:\)
• \(1s^2, 2s^2, 3s^2, 4s^2, 5s^2\) contribute \(2 × 5 = 10\) s-electrons.
Thus, the number of s-electrons is 10.
Considering Bohr’s atomic model for hydrogen atom :
(A) the energy of H atom in ground state is same as energy of He+ ion in its first excited state.
(B) the energy of H atom in ground state is same as that for Li++ ion in its second excited state.
(C) the energy of H atom in its ground state is same as that of He+ ion for its ground state.
(D) the energy of He+ ion in its first excited state is same as that for Li++ ion in its ground state.
MX is a sparingly soluble salt that follows the given solubility equilibrium at 298 K.
MX(s) $\rightleftharpoons M^{+(aq) }+ X^{-}(aq)$; $K_{sp} = 10^{-10}$
If the standard reduction potential for $M^{+}(aq) + e^{-} \rightarrow M(s)$ is $(E^{\circ}_{M^{+}/M}) = 0.79$ V, then the value of the standard reduction potential for the metal/metal insoluble salt electrode $E^{\circ}_{X^{-}/MX(s)/M}$ is ____________ mV. (nearest integer)
[Given : $\frac{2.303 RT}{F} = 0.059$ V]
An infinitely long straight wire carrying current $I$ is bent in a planar shape as shown in the diagram. The radius of the circular part is $r$. The magnetic field at the centre $O$ of the circular loop is :
