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The incorrect statement among the following is : (1) beryllium oxide is amphoteric in nature (2) solubility of sulphates of second group elements decreases down the group (3) reducing power of hydride of alkali metal decreases down the group (4) Berylium has diagonal relationship with alumunium

Short Answer

Expert verified
Statement (3) is incorrect.

Step by step solution

01

Analyze Each Statement

Carefully read through all given statements and identify key concepts related to the elements and groups in the periodic table.
02

Understand Amphoteric Nature

Statement (1) indicates that beryllium oxide is amphoteric, which means it can act as both an acid and a base. This is a correct property of beryllium oxide.
03

Examine Solubility Trend

Statement (2) mentions that the solubility of sulphates of second group elements decreases down the group. This is a true statement: solubility of sulphates, like MgSO4, CaSO4, etc., tends to decrease as you move down the alkaline earth metal group.
04

Evaluate Reducing Power of Hydrides

Statement (3) states that the reducing power of alkali metal hydrides decreases down the group. This is incorrect. In fact, the reducing power increases as you move down the group due to the larger atomic size and lower ionization energy.
05

Diagonal Relationship

Statement (4) says that beryllium has a diagonal relationship with aluminum. This is correct due to their similar properties in terms of charge/radius ratio and chemistry.
06

Identify the Incorrect Statement

From the analysis, the incorrect statement is: (3) reducing power of alkali metal hydrides decreases down the group.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

periodic table trends
The periodic table is a powerful tool for understanding the elements and their behaviors. A few key trends can help us predict how elements will react and combine. For example, atomic size generally increases as you move down a group due to the addition of electron shells. Conversely, it decreases across a period from left to right as the increasing number of protons pulls electrons closer.

The table can also predict ionization energy, electronegativity, and metallic character. Ionization energy decreases as you go down a group and increases across a period. Electronegativity follows a similar trend. These trends are essential for understanding concepts like solubility, reactivity, and bond formation.

alkali metals and hydrides
Alkali metals like lithium, sodium, and potassium are highly reactive, especially with water. They form hydrides by combining with hydrogen. These hydrides have interesting characteristics. One crucial trend is that their reducing power increases as you move down the group. This is because larger atoms lose electrons more easily due to lower ionization energies.

Alkali metal hydrides (e.g., NaH, KH) are strong reducing agents, often used in chemical reactions to donate electrons. Understanding these trends helps predict reactions and product formation in various chemical processes.
solubility of sulphates
When looking at the alkaline earth metals (Group 2 elements), we can observe an interesting trend in the solubility of their sulphates. Sulphates of these metals become less soluble as you move down the group. For example, beryllium sulphate (BeSO4) is highly soluble in water, while barium sulphate (BaSO4) is virtually insoluble.

This pattern occurs because larger metal ions have lower lattice energies and more hydration energy. As a result, the ionic bonds between the metal cation and sulphate anion become weaker, decreasing solubility.

amphoteric oxides
Some oxides, like beryllium oxide (BeO), can act as either acids or bases. These are called amphoteric oxides. They react with both acids and bases to form salts and water. Amphoteric behavior is due to the oxide's ability to donate or accept electrons based on the reacting substance.

For beryllium oxide, it can react with hydrochloric acid to form beryllium chloride (BeCl2) and water, or it can react with sodium hydroxide to form sodium beryllate (Na2BeO2) and water. This dual behavior makes amphoteric oxides interesting and versatile in chemical reactions.

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Most popular questions from this chapter

The mean and variance of a binomial distribution are 30 and 5 respectively. The probability of two successes is (1) \(\frac{875}{6^{34}}\) (2) \(\frac{875}{2.6^{34}}\) (3) \(\frac{875}{6^{36}}\) (4) \({ }^{36} \mathrm{C}_{2}\left(\frac{5}{6}\right)^{34}\left(\frac{1}{6}\right)^{2}\)

In a damped oscillator the amplitude of vibrations of mass \(\mathrm{m}=150\) grams falls by \(\frac{1}{\mathrm{e}}\) times of its initial value in time \(t_{0}\) due to viscous forces. The time \(t_{0}\) and the percentage loss in mechanical energy during the above time interval \(t_{0}\) respectively are (Let damping constant be \(50 \mathrm{grams} / \mathrm{s}\) ) (1) 6s, \(\frac{\mathrm{e}^{2}-1}{\mathrm{e}^{2}} \times 100\) (2) 3s, \(\frac{e^{2}-1}{e^{2}} \times 100\) (3) 6s, \(\frac{e-1}{e} \times 100\) (4) 3s, \(\frac{e-1}{e} \times 100\)

Which of the following statements is not correct from the point of view of molecular orbital? (1) \(\mathrm{Be}_{2}\) is not a stable molecule (2) \(\mathrm{He}_{2}\) not stable but \(\mathrm{He}^{+}\)is expected to exist (3) Bond strength of \(\mathrm{N}_{2}\) is maximum amongst the homonuclear diatomic molecules (4) The order of energies of molecular orbitals in \(\mathrm{O}_{2}\) molecule is \(\mathrm{E}(\sigma 2 \mathrm{~s})<\mathrm{E}\left(\sigma^{*} 2 \mathrm{~s}\right)<\mathrm{E}\left(\pi 2 p_{\mathrm{x}}\right)=\mathrm{E}\left(\pi 2 \mathrm{p}_{\mathrm{y}}\right)<\) \(\mathrm{E}\left(\sigma 2 \mathrm{p}_{2}\right)<\mathrm{E}\left(\pi^{\star} 2 \mathrm{p}_{\mathrm{x}}\right)=\mathrm{E}\left(\pi^{*} 2 \mathrm{p}_{y}\right)<\mathrm{E}\left(\sigma^{\star} 2 \mathrm{p}_{\mathrm{z}}\right)\)

Statement-1 : If kinetic energy of a body is constant, the net force on it is zero. Statement-2 : According to work energy theorem work done by all forces on a system converts into kinetic energy of system. (1) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1. (2) Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1 (3) Statement-1 is True, Statement-2 is False (4) Statement- 1 is False, Statement-2 is True

\(\mathrm{S}=\sum_{\mathrm{r}=1}^{7} \tan ^{2} \frac{\mathrm{r} \pi}{16}\), then \(\mathrm{S}\) is equal to (1) 44 (2) 40 (3) 34 (4) 35

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