Magnetic moment, $\mu=\sqrt{n(n+2)}$
Where, $n=$ number of unpaired electrons
The electronic configuration for the given ions is as follow:
For $M n^{2+}[Z=25]=1 s^{2}, 2 s^{2}, 2 p^{6}, 3 s^{2} 3 p^{6}, 3 d^{2} 4 s^{0}$
$\mathop {\boxed \uparrow \boxed \uparrow \boxed \uparrow \boxed \uparrow \boxed \uparrow }\limits^{3{d^2}} $
$\mu=\sqrt{5(5+2)}=\sqrt{35}$
For $C r^{2+}[Z=24]=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6}, 3 d^{4}, 4 s^{0}$
$\mathop {\boxed \uparrow \boxed \uparrow \boxed \uparrow \boxed \uparrow }\limits^{3{d^4}} $
$\mu=\sqrt{4(4+2)}=\sqrt{24}$
For $T i^{2+}[Z=22]=1 s^{2}, 2 s^{2} 2 p^{6}, 3 s^{2} 3 p^{6}, 3 d^{2}, 4 s^{0}$
$\mathop {\boxed \uparrow \boxed \uparrow \boxed \uparrow }\limits^{3{d^2}} $
$\mu=\sqrt{2(2+2)}=\sqrt{8}$
$\therefore$ correct order of spin only magnetic moment is
$M n^{2}+C r^{2+}>T i^{2+}$
સૂચિ $I$ (સવર્ગ સંયોજન સ્પીસીઝ) | સૂચિ $II$ (અવશોષિત પ્રકાશની તરંગલંબાઈ $nm$) |
$A$ ${\left[ CoCl \left( NH _3\right)_5\right]^{2+}}$ | $I$ $310$ |
$B$ ${\left[ Co \left( NH _3\right)_6\right]^{3+}}$ | $II$ $475$ |
$C$ ${\left[ Co ( CN )_6\right]^{3-}}$ | $III$ $535$ |
$D$ ${\left[ Cu \left( H _2 O \right)_4\right]^{2+}}$ | $IV$ $600$ |
નીચે આપેલા વિકલ્પોમાંથી સાચો જવાબ પસંદ કરો.
$(i)\, Pt(SCN)_2 · 3PEt_3$
$(ii)\, CoBr · SO_4 · 5NH_3$
$(iii)\, FeCl_3 · 6H_2O$
સૂચિ $- I$ | સૂચિ $- II$ |
$(A)\,Ni(CN)^{3-}_5$ | $(1)\, sp^3$ |
$(B)\, CuCl^{3-}_5$ | $(2)\, dsp^2$ |
$(C)\, AuCl^-_4$ | $(3)\, sp^3d_{z^2}$ |
$(D) \,ClO^-_4$ | $(4)\, d_{x^2-y^2} sp^3$ |
$A\,\,\,-\,\,\,B\,\,\,-\,\,\,C\,\,\,-\,\,\,D$
(પ. ક્ર.: $Sc = 21 , Ti = 22, V = 23, Zn = 30$)
$(I)\, [Ni(CN)_4]^{2-}\,\,\,(II)\, [NiCl_4]^{2-}\,\,\,(III)\,Ni(CO_4)\,\,\,$
$(IV)\, [Ni(H_2O)_6 ]^{2+}$
${\left[ Ni ( CN )_4\right]^{2-},\left[ Ni ( CO )_4\right],\left[ NiCl _4\right]^{2-}}$
${\left[ Fe ( CN )_6\right]^{4-},\left[ Cu \left( NH _3\right)_4\right]^{2+}}$
${\left[ Fe ( CN )_6\right]^{3-} \text { and }\left[ Fe \left( H_2O\right)_6\right]^{2+}}$