Question

in each pair, choose the element with the bigger atoms.
note: solve this problem using only the location of each element in the periodic table.
element pair | element with the larger atomic radius
cd | ○
ag | ○
re | ○
la | ○
nb | ○
mo | ○
ta | ○
w | ○

Explanation:

Step1: Recall atomic radius trends

Atomic radius increases down a group (more electron shells) and decreases across a period (increased nuclear charge pulls electrons closer).

Step2: Analyze each pair

• Cd vs Ag: Ag is in period 5, group 11; Cd is period 5, group 12. Across period, radius decreases left to right? Wait, no: Ag (group 11) and Cd (group 12) are adjacent. Wait, actually, Ag is to the left of Cd? No, Ag is group 11, Cd group 12. Wait, no: period 5, elements go from Rb (group 1) to Xe (group 18). So Ag (47) is group 11, Cd (48) group 12. So left to right, atomic radius decreases. Wait, but Ag is left of Cd? No, Ag is 47, Cd 48. So Ag is left, so Ag has larger radius? Wait, no, wait: atomic radius decreases across a period (left to right) because nuclear charge increases, pulling electrons in. So Ag (left) has larger radius than Cd (right) in same period. Wait, but wait, Cd is a transition metal, Ag too. Wait, maybe I messed up. Wait, let's check periods and groups.

• Re vs La: La is in period 6, group 3 (lanthanide); Re is period 6, group 7. Across period, radius decreases left to right. La is left of Re, so La has larger radius.

• Nb vs Mo: Nb (period 5, group 5), Mo (period 5, group 6). Left to right, radius decreases. So Nb (left) has larger radius.

• Ta vs W: Ta (period 6, group 5), W (period 6, group 6). Left to right, radius decreases. So Ta (left) has larger radius.

Wait, let's recheck each pair:

1. Cd (48, period 5, group 12) vs Ag (47, period 5, group 11): Same period, Ag is left of Cd. Atomic radius decreases left to right? Wait, no: in period 5, from Rb (group 1) to Xe (group 18), atomic radius decreases. So Ag (group 11) is left of Cd (group 12), so Ag has larger radius. Wait, but wait, Cd is a d-block element, Ag too. Wait, maybe I made a mistake. Wait, actually, for transition metals, the trend is less clear, but generally, across a period, atomic radius decreases. So Ag (left) > Cd (right) in period 5.

1. Re (75, period 6, group 7) vs La (57, period 6, group 3): La is in lanthanide series, which has lanthanide contraction, but La is at the start of period 6, group 3. Re is in d-block, group 7, period 6. So La is left of Re, so La has larger radius.

1. Nb (41, period 5, group 5) vs Mo (42, period 5, group 6): Same period, Nb left of Mo, so Nb has larger radius.

1. Ta (73, period 6, group 5) vs W (74, period 6, group 6): Same period, Ta left of W, so Ta has larger radius.

So the elements with larger atomic radius are:

• Ag (vs Cd)

• La (vs Re)

• Nb (vs Mo)

• Ta (vs W)

Wait, but the table has each pair:

First row: Cd and Ag → choose Ag

Second row: Re and La → choose La

Third row: Nb and Mo → choose Nb

Fourth row: Ta and W → choose Ta

So for each pair, select the left element (in same period) as atomic radius decreases left to right.

Step1: Cd vs Ag

Ag is in period 5, group 11; Cd in period 5, group 12. Same period, left (Ag) has larger radius. So select Ag.

Step2: Re vs La

La (period 6, group 3) is left of Re (period 6, group 7). Same period, left has larger radius. Select La.

Step3: Nb vs Mo

Nb (period 5, group 5) left of Mo (period 5, group 6). Same period, left has larger radius. Select Nb.

Step4: Ta vs W

Ta (period 6, group 5) left of W (period 6, group 6). Same period, left has larger radius. Select Ta.

Answer:

For each pair:

• Cd vs Ag: Ag (select the circle next to Ag)

• Re vs La: La (select the circle next to La)

• Nb vs Mo: Nb (select the circle next to Nb)

• Ta vs W: Ta (select the circle next to Ta)