For the ions Sc(.III.),Ti(.IV.),Pd(.II.) and Cu(.II.) , which of the following option is correct?

Question

For the ions Sc(.III.),Ti(.IV.),Pd(.II.) and Cu(.II.) , which of the following option is correct? (A) All are paramagnetic (B) All are diamagnetic (C) Sc(.III.),Ti(.IV.) are paramagnetic and Pd(.II.),Cu(.II.) are diamagnetic (D) Sc(.III.),Ti(.IV.) are diamagnetic and Pd(.II.),Cu(.II.) are paramagnetic

Tardigrade Admin · Accepted Answer

Sc3 + (3 d0), Ti4 +(.3d0.) are diamagnetic due to absence of unpaired electrons. While Pd2 +(4 d8), Cu2 +(.3d9.) contain two, and one unpaired electron respectively. Hence, these are paramagnetic

Q. For the ions $S c (III), T i (I V), P d (II)$ and $C u (II)$ , which of the following option is correct?

All are paramagnetic

All are diamagnetic

$S c (III), T i (I V)$ are paramagnetic and $P d (II), C u (II)$ are diamagnetic

$S c (III), T i (I V)$ are diamagnetic and $P d (II), C u (II)$ are paramagnetic

$S c^{3 +} (3 d^{0}), T i^{4 +} (3 d^{0})$ are diamagnetic due to absence of unpaired electrons. While $P d^{2 +} (4 d^{8}), C u^{2 +} (3 d^{9})$ contain two, and one unpaired electron respectively. Hence, these are paramagnetic

Q. For the ions Sc(III),Ti(IV),Pd(II) and Cu(II) , which of the following option is correct?

All are paramagnetic

All are diamagnetic

Sc(III),Ti(IV) are paramagnetic and Pd(II),Cu(II) are diamagnetic

Sc(III),Ti(IV) are diamagnetic and Pd(II),Cu(II) are paramagnetic

Sc3+(3d0),Ti4+(3d0) are diamagnetic due to absence of unpaired electrons. While Pd2+(4d8),Cu2+(3d9) contain two, and one unpaired electron respectively. Hence, these are paramagnetic

Q. For the ions $S c (III), T i (I V), P d (II)$ and $C u (II)$ , which of the following option is correct?

$S c (III), T i (I V)$ are paramagnetic and $P d (II), C u (II)$ are diamagnetic

$S c (III), T i (I V)$ are diamagnetic and $P d (II), C u (II)$ are paramagnetic

$S c^{3 +} (3 d^{0}), T i^{4 +} (3 d^{0})$ are diamagnetic due to absence of unpaired electrons. While $P d^{2 +} (4 d^{8}), C u^{2 +} (3 d^{9})$ contain two, and one unpaired electron respectively. Hence, these are paramagnetic