Q.Identify the group and valency of the element having atomic number 119. Also predict the outermost electronic configuration and write the general formula of its oxide.

The question tests your understanding of the inert-pair effect in group 14 elements. As you go down the group from carbon to lead, the stability of the +4 oxidation state decreases while the +2 oxidation state becomes more stable. This happens because the 6s electrons in lead are held very tightly to the nucleus (relativistic contraction) and resist being shared or lost in bonding. So Pb(IV) compounds are strong oxidising agents and only form with small, highly electronegative atoms like fluorine that can stabilise the high oxidation state. Large, easily polarised halides like iodide cannot do this — they get oxidised instead.

Let’s check each option one by one.

1. $\mathrm{SnF}_4$ — Tin(IV) fluoride. Tin is above lead in group 14, so the inert-pair effect is weaker. Sn(IV) is reasonably stable, and fluorine is the most electronegative element, easily stabilising the +4 state. This compound exists and is a white solid.

2. $\mathrm{PbF}_4$ — Lead(IV) fluoride. Lead(IV) is unstable, but fluorine’s tiny size and extreme electronegativity can still pull electrons away from lead to form Pb–F bonds. This compound exists, though it is a strong oxidising agent and decomposes on heating. It is known.

3. $\mathrm{PbI}_4$ — Lead(IV) iodide. Here’s the problem. Iodide is a large, easily oxidised ion ($\mathrm{I}^-$ to $\mathrm{I}_2$). Pb(IV) is such a strong oxidiser that it would immediately oxidise iodide to iodine and get reduced to Pb(II) itself. The reaction would be: $\mathrm{PbI}_4 \rightarrow \mathrm{PbI}_2 + \mathrm{I}_2$. So $\mathrm{PbI}_4$ has never been isolated — it simply does not exist under normal conditions. …

The question asks: in the long-form periodic table, an element X is present in group Y, and Mn (manganese) is also an element in that same group. We need to identify X and Y.

First, recall where Mn sits. Manganese has atomic number 25. In the long-form periodic table, it is a d-block element (transition metal) placed in the 4th period. Its group number is 7 — because the d-block groups are numbered 3 through 12, and Mn is in the 7th column of the d-block (counting from Sc, group 3). So Y must be 7. …

The problem asks us to identify three elements, X, Y, and Z, which have successive atomic numbers, based on their first ionization enthalpies. Ionization enthalpy is a fundamental periodic property, and its trends are key to solving this.

Concept and Intuition

Ionization enthalpy (IE) is the minimum energy required to remove the most loosely bound electron from an isolated gaseous atom in its ground state.

The general trends for ionization enthalpy are:

• Across a period: Ionization enthalpy generally increases from left to right. This is because the effective nuclear charge increases, pulling the valence electrons more strongly towards the nucleus, making them harder to remove.
• Down a group: Ionization enthalpy generally decreases. This is due to the increasing atomic size and greater shielding effect from inner electrons, which reduce the attraction between the nucleus and the outermost electron.

However, there are specific elements that show significant deviations from these general trends, which are crucial for this problem:

1. Noble Gases (Group 18): These elements have completely filled valence electron shells ($ns^2np^6$), which are exceptionally stable. Consequently, they have the highest ionization enthalpies in their respective periods. Removing an electron from such a stable configuration requires a very large amount of energy.
2. Alkali Metals (Group 1): These elements have only one electron in their outermost shell ($ns^1$). Removing this single electron results in a stable noble gas configuration. Therefore, alkali metals have the lowest ionization enthalpies in their respective periods.
3. Alkaline Earth Metals (Group 2): These elements have two electrons in their outermost shell ($ns^2$). Their ionization enthalpies are higher than those of alkali metals in the same period but significantly lower than those of noble gases.

The given ionization enthalpy values are 2080, 496, and 737 kJ mol$^{-1}$. Let's analyze these values:

• 2080 kJ mol$^{-1}$: This is an exceptionally high value. This immediately suggests that the element corresponding to this IE (X) must be a noble gas.
• 496 kJ mol$^{-1}$: This is a very low value. This suggests that the element corresponding to this IE (Y) must be an alkali metal.
• 737 kJ mol$^{-1}$: This value is higher than 496 but much lower than 2080. This suggests that the element corresponding to this IE (Z) is likely an alkaline earth metal.

Since X, Y, and Z have successive atomic numbers, if X is a noble gas, then Y must be the element immediately following it in the periodic table (an alkali metal in the next period), and Z must be the element following Y (an alkaline earth metal). This sequence (Noble gas $\rightarrow$ Alkali metal $\rightarrow$ Alkaline earth metal) perfectly matches the observed ionization enthalpy pattern: Very high IE $\rightarrow$ Very low IE $\rightarrow$ Moderately low IE (higher than alkali metal).

Step-by-step Solution

1. Identify the element with the highest ionization enthalpy:

   The ionization enthalpies are given as 2080, 496, and 737 kJ mol$^{-1}$. The value 2080 kJ mol$^{-1}$ is significantly higher than the other two. This indicates that the element X, which has this ionization enthalpy, possesses a very stable electron configuration, characteristic of a noble gas.

2. Relate the elements based on successive atomic numbers:

   Since X, Y, and Z have successive atomic numbers, if X is a noble gas, then Y must be the element immediately after it in the periodic table. The element immediately following a noble gas is always an alkali metal, starting a new period. The element immediately following an alkali metal is an alkaline earth metal.

   So, the sequence of elements must be: Noble Gas (X) $\rightarrow$ Alkali Metal (Y) $\rightarrow$ Alkaline Earth Metal (Z).

3. Check if the ionization enthalpy values fit this sequence:

   • X (Noble Gas): Should have a very high IE. The given value for X is 2080 kJ mol$^{-1}$, which fits this criterion.
   • Y (Alkali Metal): Should have a very low IE (the lowest in its period). The given value for Y is 496 kJ mol$^{-1}$, which is indeed very low. …

The configuration $(n-1)d^{10}ns^2$ means the penultimate shell holds a completely filled $d^{10}$ and the outermost shell holds $ns^2$. In the ground state this is characteristic of the group-12 elements.

Screening the options by ground-state valence configuration:

• (A) Bh, Eu, Po - Bh $=[\text{Rn}]5f^{14}6d^{5}7s^{2}$ ($d^5$); Eu $=[\text{Xe}]4f^{7}6s^{2}$ (no valence $d$). Rejected.
• (B) Ho, Er, Lu - all lanthanoids: Ho $4f^{11}6s^2$, Er $4f^{12}6s^2$, Lu $[\text{Xe}]4f^{14}5d^{1}6s^{2}$ ($d^1$). Rejected.
• (C) Hs, Hg, W - contains Hg $=[\text{Xe}]4f^{14}5d^{10}6s^{2}$, the exact $(n-1)d^{10}ns^2$ match.
• (D) K, Bi, Ba - K $=[\text{Ar}]4s^1$; Ba $=[\text{Xe}]6s^2$ (no valence $d$). Rejected. …

Concept & Intuition

The question asks: In Group 13, which element has the greatest difference between its boiling point and melting point?

This is not about trends in absolute values, but about the spread between the two. The key is to recall that gallium (Ga) has a famously low melting point (it melts in your hand, around 30 °C), yet its boiling point is very high (over 2400 °C). That huge gap makes it the standout. Boron, by contrast, has both a very high melting and boiling point (both > 2000 °C), so the difference is smaller. Aluminium and indium have moderate differences, but none come close to gallium’s anomaly.

Step-by-step reasoning

1. Recall the melting and boiling points of Group 13 elements (in Kelvin)

   • Boron (B): mp ≈ 2349 K, bp ≈ 4200 K → ΔT ≈ 1851 K
   • Aluminium (Al): mp ≈ 933 K, bp ≈ 2792 K → ΔT ≈ 1859 K
   • Gallium (Ga): mp ≈ 303 K, bp ≈ 2676 K → ΔT ≈ 2373 K
   • Indium (In): mp ≈ 430 K, bp ≈ 2345 K → ΔT ≈ 1915 K

   (These are approximate standard values; small variations exist in different sources, but the relative sizes are consistent.)

2. Identify the anomaly

   Gallium’s melting point is dramatically lower than the others — it is only about 30 °C above room temperature. This is due to its unique crystal structure (orthorhombic) with weak interatomic bonding in one direction, making it melt easily. Its boiling point, however, is still high because breaking all metallic bonds requires much more energy.

3. Compare the differences

   • B: ~1850 K
   • Al: ~1860 K
   • Ga: ~2370 K
   • In: ~1915 K …

Zeolites are a fascinating class of minerals with a unique structure that gives them remarkable properties, making them invaluable in various industrial applications, from water purification to catalysis. Understanding their composition is key to appreciating their function.

The term "silicate" refers to compounds containing silicon and oxygen, often in a tetrahedral arrangement ($\text{SiO}_4^{4-}$). Zeolites are not just silicates; they are specifically aluminosilicates. This means that in their crystal structure, some of the silicon atoms in the silicate framework are replaced by aluminium atoms.

1. Understanding the Zeolite Framework:

   The fundamental building blocks of a zeolite structure are $\text{SiO}_4$ and $\text{AlO}_4$ tetrahedra. These tetrahedra link together by sharing oxygen atoms to form a three-dimensional, open framework with pores and channels.

2. The Role of Aluminium (Al):

   When an $\text{Al}^{3+}$ ion replaces an $\text{Si}^{4+}$ ion in the tetrahedral framework, it introduces a net negative charge. This is because $\text{Si}^{4+}$ has a charge of $+4$, while $\text{Al}^{3+}$ has a charge of $+3$. The oxygen atoms shared in the framework maintain their valency, but the overall charge of the $\text{AlO}_4$ tetrahedron becomes $-5$ (one $\text{Al}^{3+}$ and four $\text{O}^{2-}$), compared to $-4$ for $\text{SiO}_4$ (one $\text{Si}^{4+}$ and four $\text{O}^{2-}$). This charge imbalance must be compensated.

3. The Role of Charge-Balancing Cations (e.g., Na):

   To maintain electrical neutrality in the overall structure, extra-framework cations are present within the pores and channels of the zeolite. These cations balance the negative charge introduced by the aluminium substitution. Common charge-balancing cations include alkali metal ions like $\text{Na}^+$ and $\text{K}^+$, and alkaline earth metal ions like $\text{Ca}^{2+}$ and $\text{Mg}^{2+}$. These cations are typically mobile and can be exchanged, which is why zeolites are often used as ion exchangers.

   Important

   The presence of aluminium in the silicate framework is a defining characteristic of zeolites. Without aluminium, the structure would simply be a pure silica polymorph (like quartz), which does not have the same ion-exchange or catalytic properties.

4. Identifying X and Y:

   The question asks for two elements X and Y that form the silicate. Based on the explanation above: …

Why temporary names existed

Before the 1990s, newly discovered superheavy elements often had competing claims from different laboratories, and their permanent names were disputed for years. To avoid confusion, IUPAC introduced a systematic temporary naming scheme based purely on the atomic number. Each digit was assigned a Latin root:

The roots were strung together, ending in -ium, and the symbol was formed from the first letter of each root.

For element 104: $1$-$0$-$4$ → un-nil-quad-ium → Unq.

In 1997, IUPAC officially named element 104 Rutherfordium (symbol Rf) after Ernest Rutherford.

Electronic configuration of element 104

Element 104 sits in the 7th period, Group 4 of the periodic table, directly below Hafnium (Hf, element 72). It is the first of the transactinide elements, meaning it comes immediately after the actinide series ends at Lawrencium (Lr, element 103).

1. Core: The nearest noble gas is Radon (Rn, element 86), so we start with [Rn].

2. Filling the 5f subshell: The actinide series (elements 90–103) fills the $5f$ orbitals. By element 103 (Lr), all fourteen $5f$ electrons are in place: $5f^{14}$. …

The periodic table is a vast collection of elements, but only a handful exist as gases under normal conditions (room temperature and atmospheric pressure). This is a classic memory-based question in chemistry, but it’s rooted in the physical states of elements — determined by their intermolecular forces and atomic structure. Gases are typically non-metals with low boiling points, like hydrogen, nitrogen, and the noble gases.

To answer accurately, you need to recall the specific elements that are gaseous at standard conditions. Let’s list them systematically.

1. The diatomic gases: These are elements that form diatomic molecules and are gases at room temperature. They are:

   • Hydrogen ($\text{H}_2$)
   • Nitrogen ($\text{N}_2$)
   • Oxygen ($\text{O}_2$)
   • Fluorine ($\text{F}_2$)
   • Chlorine ($\text{Cl}_2$) That’s 5 elements.

2. The noble gases: These are monatomic gases, all of which are gaseous under normal conditions. They are:

   • Helium ($\text{He}$)
   • Neon ($\text{Ne}$)
   • Argon ($\text{Ar}$)
   • Krypton ($\text{Kr}$)
   • Xenon ($\text{Xe}$)
   • Radon ($\text{Rn}$) That’s 6 elements. …