Answer
Verified
375.9k+ views
Hint:If any complex has more unpaired electrons means it has a high spin, while less unpaired electron results in low spin. Inner orbital complexes contain the central atom using inner d orbital, while outer orbital complexes use outer d orbitals in the central metal.
Complete step-by-step answer:We have been given a complex ${{[Mn{{(CN)}_{6}}]}^{4-}}$, Mn has a configuration $M{{n}^{2+}}[Ar]3{{d}^{5}}$ , with CN being a ligand, this complex consist of unpaired electrons, hence it is paramagnetic, while it is an inner orbital complex with low spin due to the nature of the ligand CN. So, now we will take out its magnetic moment.
Magnetic moment of any complex is its behavior due to an unpaired electron under the influence of a magnetic field. It is known as a spin only magnetic moment as it is based on the spin of the unpaired electron. The formula for calculation of magnetic moment is, ${{\mu }_{s}}=\sqrt{n(n+2)BM}$,where ${{\mu }_{s}}$, is the spin magnetic moment, n is number of unpaired electron, BM is known as Bohr magneton.
Now, taking out the magnetic moment, we have n = 1 for this complex, so,
${{\mu }_{s}}=\sqrt{1(1+2)BM}$
${{\mu }_{s}}=\sqrt{3BM}$
${{\mu }_{s}}$= 1.73 BM
So, the magnetic moment is 1.73 BM, hence, option C is correct for this complex.
Note:
Outer orbital complexes have the hybridization of dsp order, while inner orbital complexes have the hybridization of spd order. The nature of ligands lead to paramagnetic or diamagnetic behavior in coordination complexes.
Complete step-by-step answer:We have been given a complex ${{[Mn{{(CN)}_{6}}]}^{4-}}$, Mn has a configuration $M{{n}^{2+}}[Ar]3{{d}^{5}}$ , with CN being a ligand, this complex consist of unpaired electrons, hence it is paramagnetic, while it is an inner orbital complex with low spin due to the nature of the ligand CN. So, now we will take out its magnetic moment.
Magnetic moment of any complex is its behavior due to an unpaired electron under the influence of a magnetic field. It is known as a spin only magnetic moment as it is based on the spin of the unpaired electron. The formula for calculation of magnetic moment is, ${{\mu }_{s}}=\sqrt{n(n+2)BM}$,where ${{\mu }_{s}}$, is the spin magnetic moment, n is number of unpaired electron, BM is known as Bohr magneton.
Now, taking out the magnetic moment, we have n = 1 for this complex, so,
${{\mu }_{s}}=\sqrt{1(1+2)BM}$
${{\mu }_{s}}=\sqrt{3BM}$
${{\mu }_{s}}$= 1.73 BM
So, the magnetic moment is 1.73 BM, hence, option C is correct for this complex.
Note:
Outer orbital complexes have the hybridization of dsp order, while inner orbital complexes have the hybridization of spd order. The nature of ligands lead to paramagnetic or diamagnetic behavior in coordination complexes.
Recently Updated Pages
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
Mark and label the given geoinformation on the outline class 11 social science CBSE
When people say No pun intended what does that mea class 8 english CBSE
Name the states which share their boundary with Indias class 9 social science CBSE
Give an account of the Northern Plains of India class 9 social science CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
In Indian rupees 1 trillion is equal to how many c class 8 maths CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Why is there a time difference of about 5 hours between class 10 social science CBSE