Classification Of Elements And Periodicity In Properties Chapter-Wise Test 1

Correct answer Carries: 4.

Wrong Answer Carries: -1.

Which of the following elements exhibits a diagonal relationship with Be?

Be (group 2, period 2) shows a diagonal relationship with Al (group 13, period 3) due to similar size and electronegativity, unlike Mg, Ca, or B.

Mg
Ca
Al
B
3

Which group 16 element forms a hydride with the lowest boiling point?

Boiling points of group 16 hydrides increase with molecular weight, except for \(\ce{H2O}\) (hydrogen bonding). \(\ce{H2S}\) has the lowest boiling point among \(\ce{H2O}\), \(\ce{H2S}\), \(\ce{H2Se}\), and \(\ce{H2Te}\) due to weak intermolecular forces.

\(\ce{H2O}\)
\(\ce{H2Se}\)
\(\ce{H2Te}\)
\(\ce{H2S}\)
4

Which element in group 16 has the lowest first ionization enthalpy?

First ionization enthalpy decreases down a group due to increasing size. Among O, S, Se, and Te, Te (period 5) has the lowest value.

O
S
Se
Te
4

Which of the following is a metalloid?

Metalloids lie along the metal-non-metal boundary. Si (group 14) is a metalloid, while Na (group 1), Mg (group 2), and S (group 16) are not.

Na
Mg
Si
S
3

Which element in the fourth period has the highest third ionization enthalpy?

Third ionization enthalpy peaks when removing an electron from a noble gas-like \(\ce{X^2+}\) ion. Ca (after losing two electrons, becomes \(\ce{Ca^2+}\) with \(1s^2 2s^2 2p^6 3s^2 3p^6\)) has the highest value.

K
Sc
Ca
Ti
3

Which element in group 2 has the highest second ionization enthalpy?

Second ionization enthalpy for group 2 elements involves removing an electron from \(\ce{X+}\) (noble gas-like). Be⁺ (smallest size) has the highest value due to greater nuclear attraction.

Mg
Ca
Sr
Be
4

Which element was predicted as Eka-silicon by Mendeleev?

Mendeleev predicted Eka-silicon below Si, later identified as Germanium (Ge), matching its properties like oxide formula \(\ce{GeO2}\).

Ga
In
Sn
Ge
4

Which of the following is an s-block element?

s-block elements have outermost electrons in s-orbitals (groups 1 and 2). Na (group 1) is an s-block element, while B (group 13), Si (group 14), and P (group 15) are p-block elements.

Na
B
Si
P
1

Arrange the following elements in order of decreasing second ionization enthalpy: Li, Be, B, C.

Second ionization enthalpy is highest for Li (removing e⁻ from \(\ce{Li+}\) with \(1s^2\)) and decreases across period 2 as stability reduces: Li > Be > B > C.

C > B > Be > Li
Li > Be > B > C
Be > Li > C > B
B > C > Li > Be
2

Which element in the fifth period has the highest sixth ionization enthalpy?

Sixth ionization enthalpy peaks when removing an electron from a stable \(\ce{X^5+}\) ion. Te (after losing five electrons, becomes \(\ce{Te^5+}\) with \(1s^2 2s^2 2p^6 3s^2 3p^6 3d^{10} 4s^2 4p^6\)) has the highest value in period 5.

Te
Sb
Sn
I
1

Performance Summary

Score:

Category Details
Total Attempts:
Total Skipped:
Total Wrong Answers:
Total Correct Answers:
Time Taken:
Average Time Taken per Question:
Accuracy:
0
error: Content is protected !!