Consider transition metal ions which have d1 to d10 electrons. Which of these can form high spin and low spin complexes in octahedral geometry.
Correct answer: B. d4, d5, d6, d7
- A. d3, d5, d6, d9
- B. d4, d5, d6, d7
- C. d2, d5, d7, d9
- D. d4, d6, d7, d10
Explanation
Metal ions with d4, d5, d6 and d7 configuration can form high spin and low spin complexes. For transition metal ions with configuration d1, d2, d3, d8, d9 or d10 all the octahedral complexes of a given metal ion have approximately the same magnetic moment and there is no high spin versus low spin complexes.
Last updated
About Transition Elements
Transition elements are identified by their partially filled d subshells in atoms or common ions. The topic covers the electronic structure of d-block elements, including the filling order of 4s and 3d orbitals, variable oxidation states and the distinction between transition elements and elements with completely filled d subshells.
Practise Transition Elements
531 free Transition Elements MCQs from Chemistry, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Exams that ask Chemistry questions like this
Chemistry is on 14 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
Related questions
A cobalt-amine has the formula Co(NH3)6Cl3. With AgNO3 solution, one third of the chloride is precipitated. The complex shows cis-trans isomerism. It can have the structure:
A complex compound of cobalt has a molecular formula containing five NH3 molecules, one nitro group and two chlorine atoms for one Co atom. One mole of this compound produces three moles of ions in an aqueous solution. On reacting with excess AgNO3 solution, two moles of AgCl get precipitated. The ionic formula of the compound is?
A compound having coordination number 6 has geometry:
A coordination complex compound of cobalt has a molecular formula containing five ammonia molecules, one nitro group, and two chlorine atoms for one cobalt atom. One mole of this compound produces three-mole ions in an aqueous solution. By the reaction of this solution with excess silver nitrate solution, two molecules of AgCl get precipitated. The ionic formula of this compound would be:
Acidified K₂Cr₂O₇ is also called: