Question

Which of the following is / are incorrectly matched.

(i) Be$^{+2}$ > He > Br $\rightarrow$ IE$_{1}$ order (ii) Mt (Z = 109) element belongs to $\rightarrow$ 7$^{th}$ period 11$^{th}$ group (iii) Li$^{+}$ > Mg$^{2+}$ > Al$^{3+}$ > Be$^{2+}$ $\rightarrow$ Ionic radius order (iv) Ga > Al > Tl $\rightarrow$ IE$_{1}$ order (v) S > F $\rightarrow$ EA order (vi) CrO < Cr$_{2}$O$_{3}$ < CrO$_{3}$ $\rightarrow$ Acidic strength order (vii) Li$^{+}_{(aq.)}$ < Na$^{+}_{(aq.)}$ < K$^{+}_{(aq.)}$ $\rightarrow$ Ionic mobility order

(viii) $\frac{Na^{\ominus}}{Na}$ < $\frac{Li^{\ominus}}{Li}$ < $\frac{H^{\ominus}}{H}$ $\rightarrow$ Size ratio are

• A. (i) Be$^{+2}$ > He > Br $\rightarrow$ IE$_{1}$ order
• B. (ii) Mt (Z = 109) element belongs to $\rightarrow$ 7$^{th}$ period 11$^{th}$ group
• C. (iii) Li$^{+}$ > Mg$^{2+}$ > Al$^{3+}$ > Be$^{2+}$ $\rightarrow$ Ionic radius order
• D. (iv) Ga > Al > Tl $\rightarrow$ IE$_{1}$ order
• E. (v) S > F $\rightarrow$ EA order
• F. (vi) CrO < Cr$_{2}$O$_{3}$ < CrO$_{3}$ $\rightarrow$ Acidic strength order
• G. (vii) Li$^{+}_{(aq.)}$ < Na$^{+}_{(aq.)}$ < K$^{+}_{(aq.)}$ $\rightarrow$ Ionic mobility order
• H. (viii) $\frac{Na^{\ominus}}{Na}$ < $\frac{Li^{\ominus}}{Li}$ < $\frac{H^{\ominus}}{H}$ $\rightarrow$ Size ratio are

Answer 1

Answer: (B), (C), (D), (E)

(ii), (iii), (iv), (v)

Answer 2

The question asks us to identify the incorrectly matched statements. Let's analyze each statement:

(i) Be$^{+2}$ > He > Br $\rightarrow$ IE$_{1}$ order

This refers to the first ionization energy (IE$_{1}$) of the given species.

• IE of Be$^{+2}$: This is the energy required to remove an electron from Be$^{+2}$ (Be$^{+2}$ $\rightarrow$ Be$^{+3}$ + e$^{-}$). Be$^{+2}$ has an electron configuration of 1s$^2$. Removing an electron from a 1s orbital of a +2 ion is very difficult. This corresponds to the third ionization energy of Be (IE$_{3}$(Be)). IE$_{3}$(Be) $\approx$ 14850 kJ/mol.
• IE of He: This is the first ionization energy of Helium (He $\rightarrow$ He$^{+}$ + e$^{-}$). He has an electron configuration of 1s$^2$. IE$_{1}$(He) $\approx$ 2372 kJ/mol.
• IE of Br: This is the first ionization energy of Bromine (Br $\rightarrow$ Br$^{+}$ + e$^{-}$). Br has an electron configuration ending in 4p$^5$. IE$_{1}$(Br) $\approx$ 1140 kJ/mol.

Comparing the values: 14850 > 2372 > 1140.

Thus, Be$^{+2}$ > He > Br is the correct order for the ionization energy of these species. Statement (i) is Correct.

(ii) Mt (Z = 109) element belongs to $\rightarrow$ 7$^{th}$ period 11$^{th}$ group

Meitnerium (Mt) with atomic number Z = 109 is a synthetic element.

• Period: Elements with Z > 86 (Radon) belong to the 7th period. So, Mt belongs to the 7th period.
• Group: Mt is a d-block element. Its electron configuration is expected to be [Rn] 5f$^{14}$ 6d$^7$ 7s$^2$. For d-block elements, the group number is (number of electrons in ns orbital + number of electrons in (n-1)d orbital). Here, 2 (from 7s) + 7 (from 6d) = 9. So, Mt belongs to Group 9.

The statement claims it belongs to the 11th group, which is incorrect. Group 11 elements are Cu, Ag, Au, and Roentgenium (Rg, Z=111). Statement (ii) is Incorrect.

(iii) Li$^{+}$ > Mg$^{2+}$ > Al$^{3+}$ > Be$^{2+}$ $\rightarrow$ Ionic radius order

These are isoelectronic ions, all having 2 electrons (like Helium). For isoelectronic species, the ionic radius decreases as the nuclear charge (Z) increases because the electrons are more strongly attracted to the nucleus.

• Li$^{+}$: Z = 3 (1s$^2$)
• Be$^{2+}$: Z = 4 (1s$^2$)
• Mg$^{2+}$: Z = 12 (1s$^2$ 2s$^2$ 2p$^6$) - Wait, Mg$^{2+}$ and Al$^{3+}$ are not isoelectronic with Li$^{+}$ and Be$^{2+}$. Mg$^{2+}$ and Al$^{3+}$ are isoelectronic with Neon (10 electrons).

Let's re-evaluate.

• Li$^{+}$ (2 electrons, 1s$^2$)
• Mg$^{2+}$ (10 electrons, 1s$^2$2s$^2$2p$^6$)
• Al$^{3+}$ (10 electrons, 1s$^2$2s$^2$2p$^6$)
• Be$^{2+}$ (2 electrons, 1s$^2$)

The statement mixes two different sets of isoelectronic ions. Comparing Li$^{+}$ and Be$^{2+}$ (both 2-electron species): Z(Li) = 3, Z(Be) = 4. So, Li$^{+}$ > Be$^{2+}$ in size. Comparing Mg$^{2+}$ and Al$^{3+}$ (both 10-electron species): Z(Mg) = 12, Z(Al) = 13. So, Mg$^{2+}$ > Al$^{3+}$ in size.

Now, comparing 2-electron species with 10-electron species. 2-electron species (Li$^{+}$, Be$^{2+}$) have electrons in the 1st shell, while 10-electron species (Mg$^{2+}$, Al$^{3+}$) have electrons in the 2nd shell. Ions with electrons in higher shells are generally much larger.

So, Mg$^{2+}$ and Al$^{3+}$ should be much larger than Li$^{+}$ and Be$^{2+}$.

The correct order of ionic radii for these specific ions would be: Mg$^{2+}$ > Al$^{3+}$ > Li$^{+}$ > Be$^{2+}$.

The given order Li$^{+}$ > Mg$^{2+}$ > Al$^{3+}$ > Be$^{2+}$ is completely incorrect. Statement (iii) is Incorrect.

(iv) Ga > Al > Tl $\rightarrow$ IE$_{1}$ order

These are elements of Group 13: Aluminum (Al), Gallium (Ga), Thallium (Tl). Generally, ionization energy decreases down a group. However, there are anomalies in Group 13 due to the poor shielding effect of d- and f-electrons.

• Al (Z=13): [Ne] 3s$^2$ 3p$^1$. IE$_{1}$ = 577 kJ/mol.
• Ga (Z=31): [Ar] 3d$^{10}$ 4s$^2$ 4p$^1$. The presence of 10 3d electrons, which provide poor shielding, leads to an increase in effective nuclear charge, making Ga's IE$_{1}$ slightly higher than Al's. IE$_{1}$ = 579 kJ/mol. So, Ga > Al is correct.
• Tl (Z=81): [Xe] 4f$^{14}$ 5d$^{10}$ 6s$^2$ 6p$^1$. The presence of 14 4f and 10 5d electrons, both providing very poor shielding, leads to a significant increase in effective nuclear charge, making Tl's IE$_{1}$ higher than In and even Ga and Al. IE$_{1}$ = 589 kJ/mol.

The actual order of IE$_{1}$ for these elements is Tl (589) > Ga (579) > Al (577).

The given order Ga > Al > Tl is incorrect because Tl should be greater than Al. Statement (iv) is Incorrect.

(v) S > F $\rightarrow$ EA order

Electron Affinity (EA) is the energy released when an electron is added to a neutral gaseous atom. A more negative value indicates a higher electron affinity.

• EA of Sulfur (S): -200 kJ/mol.
• EA of Fluorine (F): -328 kJ/mol.

Since -328 kJ/mol is more negative than -200 kJ/mol, Fluorine has a higher electron affinity than Sulfur.

So, F > S in terms of electron affinity. The statement S > F is incorrect. Statement (v) is Incorrect.

(vi) CrO < Cr$_{2}$O$_{3}$ < CrO$_{3}$ $\rightarrow$ Acidic strength order

This statement compares the acidic strength of oxides of Chromium in different oxidation states.

• CrO: Chromium is in +2 oxidation state. Oxides of metals in lower oxidation states are generally basic.
• Cr$_{2}$O$_{3}$: Chromium is in +3 oxidation state. Oxides of metals in intermediate oxidation states are generally amphoteric.
• CrO$_{3}$: Chromium is in +6 oxidation state. Oxides of metals in higher oxidation states are generally acidic.

As the oxidation state of the central metal atom increases, its electronegativity increases, and the covalent character of the M-O bond increases, leading to an increase in acidic character.

Therefore, the acidic strength order is CrO (basic) < Cr$_{2}$O$_{3}$ (amphoteric) < CrO$_{3}$ (acidic). Statement (vi) is Correct.

(vii) Li$^{+}_{(aq.)}$ < Na$^{+}_{(aq.)}$ < K$^{+}_{(aq.)}$ $\rightarrow$ Ionic mobility order

Ionic mobility in aqueous solution depends on the effective size of the hydrated ion.

• Smaller ions (in gas phase) have a higher charge density and thus attract more water molecules, leading to a larger hydrated radius.
• Order of ionic size (gas phase): Li$^{+}$ < Na$^{+}$ < K$^{+}$.
• Order of hydration: Li$^{+}$ (most hydrated) > Na$^{+}$ > K$^{+}$ (least hydrated).
• Order of hydrated radius: Li$^{+}_{(aq.)}$ (largest) > Na$^{+}_{(aq.)}$ > K$^{+}_{(aq.)}$ (smallest).
• Ionic mobility is inversely proportional to the hydrated radius (larger hydrated ions move slower).

Therefore, the order of ionic mobility is Li$^{+}_{(aq.)}$ < Na$^{+}_{(aq.)}$ < K$^{+}_{(aq.)}$. Statement (vii) is Correct.

(viii) $\frac{Na^{\ominus}}{Na}$ < $\frac{Li^{\ominus}}{Li}$ < $\frac{H^{\ominus}}{H}$ $\rightarrow$ Size ratio are

This statement compares the ratio of the radius of the anion to the radius of the neutral atom (r$_{\text{anion}}$/r$_{\text{atom}}$).

When an atom gains an electron to form an anion, its size increases due to increased electron-electron repulsion and a decrease in effective nuclear charge per electron.

Let's look at approximate atomic and ionic radii (in pm):

• H: r(H) $\approx$ 37 pm, r(H$^{-}$) $\approx$ 134 pm. Ratio H$^{-}$/H = 134/37 $\approx$ 3.62.
• Li: r(Li) $\approx$ 152 pm, r(Li$^{-}$) $\approx$ 167 pm. Ratio Li$^{-}$/Li = 167/152 $\approx$ 1.10.
• Na: r(Na) $\approx$ 186 pm, r(Na$^{-}$) $\approx$ 201 pm. Ratio Na$^{-}$/Na = 201/186 $\approx$ 1.08.

Comparing the ratios: 1.08 (Na$^{-}$/Na) < 1.10 (Li$^{-}$/Li) < 3.62 (H$^{-}$/H).

The given order $\frac{Na^{\ominus}}{Na}$ < $\frac{Li^{\ominus}}{Li}$ < $\frac{H^{\ominus}}{H}$ is correct.

The large increase for H is because the electron is added to the already small 1s orbital, causing significant repulsion and expansion. For alkali metals, the electron is added to a new, larger orbital (e.g., 2s for Li, 3s for Na) which is already occupied by one electron, but the relative expansion is smaller. Statement (viii) is Correct.

Summary of Incorrect Statements: (ii), (iii), (iv), (v)

The final answer is $\boxed{\text{(ii), (iii), (iv), (v)}}$