401. A group \(1\) element \(M\) forms a simple hydride in which hydrogen behaves as \(\mathrm{H^-}\). The expected formula is
ⓐ. \(\mathrm{MH_2}\)
ⓑ. \(\mathrm{MH_3}\)
ⓒ. \(\mathrm{M_2H_3}\)
ⓓ. \(\mathrm{MH}\)
Correct Answer: \(\mathrm{MH}\)
Explanation: Group \(1\) elements commonly show oxidation state \(+1\). In ionic hydrides of very electropositive metals, hydrogen is treated as hydride ion, \(\mathrm{H^-}\). One \(M^+\) balances one \(\mathrm{H^-}\), giving the simplest formula \(\mathrm{MH}\). This formula pattern is a direct result of charge balance and group valency. It should not be confused with covalent hydrides of non-metals, where hydrogen is usually written first, as in \(\mathrm{HCl}\) or \(\mathrm{H_2S}\).
402. A group \(2\) element \(M\) forms a simple hydride with hydrogen as \(\mathrm{H^-}\). The formula that satisfies charge balance is
ⓐ. \(\mathrm{MH}\)
ⓑ. \(\mathrm{M_2H}\)
ⓒ. \(\mathrm{MH_2}\)
ⓓ. \(\mathrm{M_2H_3}\)
Correct Answer: \(\mathrm{MH_2}\)
Explanation: Group \(2\) metals commonly form \(M^{2+}\) ions by losing two valence electrons. In their ionic hydrides, hydrogen is present as \(\mathrm{H^-}\). Two hydride ions are needed to balance one \(M^{2+}\). The resulting formula is \(\mathrm{MH_2}\). The subscript \(2\) belongs to hydrogen because two negative charges are required to balance the \(+2\) metal ion.
403. The table shows hydride formula patterns for representative groups. Select the row that needs correction.
| Row | Group | Typical simple hydride formula pattern |
|---|
| P | \(1\) | \(\mathrm{MH}\) |
| Q | \(2\) | \(\mathrm{MH_2}\) |
| R | \(14\) | \(\mathrm{MH_4}\) or \(\mathrm{EH_4}\)-type covalent hydride |
| S | \(17\) | \(\mathrm{H_2E}\) |
ⓐ. Row P
ⓑ. Row Q
ⓒ. Row S
ⓓ. Row R
Correct Answer: Row S
Explanation: Group \(17\) elements have seven valence electrons and commonly show valency \(1\) in simple hydrides. Their hydrogen halides have the general formula \(\mathrm{HX}\), such as \(\mathrm{HCl}\). Row S gives \(\mathrm{H_2E}\), which is more suitable for group \(16\)-type hydrides such as \(\mathrm{H_2S}\). Rows P and Q correctly represent ionic hydride patterns of group \(1\) and group \(2\) metals. Row R is also reasonable for group \(14\), where tetravalent hydrides such as \(\mathrm{CH_4}\) are common examples.
404. Across a representative period, simple hydride formulae can change as \(\mathrm{MH}\), \(\mathrm{MH_2}\), \(\mathrm{MH_3}\), \(\mathrm{MH_4}\), \(\mathrm{H_3M}\), \(\mathrm{H_2M}\), and \(\mathrm{HM}\). The main reason for this pattern is the change in
ⓐ. neutron number only
ⓑ. atomic mass unit used for measurement
ⓒ. discovery date of elements
ⓓ. valency or combining capacity across the period
Correct Answer: valency or combining capacity across the period
Explanation: Hydride formulae reflect how many hydrogen atoms combine with one atom of the element. In the representative elements, this is closely connected with valence electrons and simple valency. On the left side, metals form hydrides such as \(\mathrm{MH}\) and \(\mathrm{MH_2}\). Toward the non-metal side, covalent hydrides such as \(\mathrm{NH_3}\), \(\mathrm{H_2O}\), and \(\mathrm{HCl}\)-type formulae appear. The formula pattern is therefore a chemical expression of periodic valency trends.
405. A representative element \(E\) forms a simple hydride \(\mathrm{EH_4}\). If hydrogen is treated as monovalent, the simplest valency of \(E\) in this hydride is
ⓐ. \(1\)
ⓑ. \(2\)
ⓒ. \(3\)
ⓓ. \(4\)
Correct Answer: \(4\)
Explanation: \( \textbf{Formula given:} \) \(\mathrm{EH_4}\).
\( \textbf{Valency of hydrogen in simple hydrides:} \) \(1\).
\( \textbf{Number of hydrogen atoms:} \) \(4\).
\( \textbf{Total combining capacity of hydrogen atoms:} \)
\[
4 \times 1=4
\]
\( \textbf{Combining capacity of one \(E\) atom:} \) One \(E\) atom combines with four monovalent hydrogen atoms.
\( \textbf{Final answer:} \) The valency of \(E\) is \(4\). The subscript in the hydride formula is a formula clue, not an atomic-number clue.
406. An element forms a hydride \(\mathrm{H_2E}\) and an oxide \(\mathrm{EO_2}\) in one common positive oxidation state. The formula pair most strongly suggests that \(E\) can behave like a
ⓐ. group \(16\) element in the hydride and a \(+4\) element in the oxide
ⓑ. group \(1\) alkali metal in both compounds
ⓒ. noble gas with valency \(0\)
ⓓ. group \(2\) metal forming only \(\mathrm{E^+}\)
Correct Answer: group \(16\) element in the hydride and a \(+4\) element in the oxide
Explanation: The hydride \(\mathrm{H_2E}\) is a common pattern for group \(16\)-type elements, where the element has simple valency \(2\). In the oxide \(\mathrm{EO_2}\), oxygen is usually \(-2\), so two oxygen atoms contribute \(-4\). This means \(E\) is assigned \(+4\) in that oxide. Many \(p\)-block elements can show different oxidation states depending on the compound. The pair shows why hydride formula and oxide formula must be interpreted with bonding context.
407. In Mendeleev-style classification, oxide and hydride formulae were useful because they gave evidence about
ⓐ. exact number of isotopes of every element
ⓑ. alphabetical order of element names
ⓒ. physical state of every element at all temperatures
ⓓ. valency and chemical similarity among elements
Correct Answer: valency and chemical similarity among elements
Explanation: Mendeleev compared formulae of oxides and hydrides to recognize chemically similar elements. Similar formula patterns often indicate similar valency or combining capacity. Elements forming related oxides and hydrides could therefore be placed in the same group. This made the table a chemical classification, not just a list of atomic masses. Formulae such as \(\mathrm{R_2O}\), \(\mathrm{RH}\), \(\mathrm{RO}\), or \(\mathrm{RH_2}\) act as clues to periodic family behaviour.
408. A group of representative elements forms hydrides of the type \(\mathrm{EH_3}\) and oxides of the type \(\mathrm{E_2O_5}\). This formula pattern most strongly suggests that the elements belong to
ⓐ. group \(1\)
ⓑ. group \(15\)
ⓒ. group \(17\)
ⓓ. group \(18\)
Correct Answer: group \(15\)
Explanation: The general hydride formula \(\mathrm{EH_3}\) shows that the element commonly has valency \(3\) toward hydrogen in this representative-group pattern. The oxide formula \(\mathrm{E_2O_5}\) shows that the element can show a \(+5\) oxidation state with oxygen. These two formula patterns are characteristic of group \(15\) elements such as nitrogen and phosphorus. Group \(1\) elements usually form \(\mathrm{MH}\)-type hydrides, group \(17\) elements form \(\mathrm{HM}\)-type hydrides, and group \(18\) elements generally have closed-shell noble-gas behaviour. Therefore, the correct group is group \(15\).
409. The oxide formula \(\mathrm{R_2O_7}\), with oxygen as \(-2\), indicates that the oxidation state of \(R\) is
ⓐ. \(+7\)
ⓑ. \(+1\)
ⓒ. \(+3\)
ⓓ. \(+5\)
Correct Answer: \(+7\)
Explanation: \( \textbf{Formula:} \) \(\mathrm{R_2O_7}\).
\( \textbf{Usual oxygen oxidation state:} \) \(-2\).
\( \textbf{Total oxygen contribution:} \)
\[
7(-2)=-14
\]
\( \textbf{Let oxidation state of \(R\) be \(x\):} \)
\[
2x-14=0
\]
\[
2x=14
\]
\[
x=+7
\]
\( \textbf{Final answer:} \) The oxidation state of \(R\) is \(+7\). This oxide pattern is typical of high positive oxidation states in the halogen region, not of group \(1\) metals.
410. A claim says, “If two elements form hydrides with the same formula type, they must have identical atomic masses.” The best evaluation is that the claim is
ⓐ. strong because formula type fixes atomic mass exactly
ⓑ. weak because similar hydride formulae suggest similar valency, not identical mass
ⓒ. strong because hydride formulae are based only on neutron number
ⓓ. weak only because hydrogen has no valency
Correct Answer: weak because similar hydride formulae suggest similar valency, not identical mass
Explanation: A hydride formula shows how an element combines with hydrogen. Similar formulae can indicate similar valency or group behaviour. However, elements in the same group usually have different atomic masses and different numbers of occupied shells. Mendeleev used formula similarities to support chemical grouping, not to claim mass equality. Formula type is a chemical-property clue rather than a direct measurement of atomic mass.
411. Consider the statements about hydrides and oxides as periodic indicators.
I. Hydride formulae can reflect valency patterns.
II. Oxide formulae can help infer oxidation state.
III. Similar oxide and hydride formulae were useful in Mendeleev's grouping.
IV. Hydride formulae always prove that two elements are isotopes.
The supported statements are
ⓐ. I and IV only
ⓑ. I, II, and III only
ⓒ. II and IV only
ⓓ. I, II, III, and IV
Correct Answer: I, II, and III only
Explanation: Statement I is correct because the number of hydrogen atoms in a simple hydride often reflects combining capacity. Statement II is correct because oxide formulae can be used with oxygen's usual oxidation state to find the element's oxidation state. Statement III is also correct because Mendeleev used oxide and hydride formulae as chemical evidence for grouping elements. Statement IV is wrong because isotopes are atoms of the same element with different mass numbers, not elements with similar hydride formulae. Formula patterns show chemical similarity, not isotope identity.
412. A periodic-table note gives these formula patterns: group \(15\), \(\mathrm{H_3E}\); group \(16\), \(\mathrm{H_2E}\); group \(17\), \(\mathrm{HE}\). The decrease in the number of hydrogen atoms from group \(15\) to group \(17\) is mainly due to
ⓐ. increasing number of neutrons in hydrogen
ⓑ. loss of all valence electrons by hydrogen
ⓒ. decreasing number of electrons needed to complete the octet
ⓓ. conversion of all elements into metals
Correct Answer: decreasing number of electrons needed to complete the octet
Explanation: Group \(15\) elements have \(5\) valence electrons and need \(3\) more electrons to complete an octet, giving hydrides such as \(\mathrm{NH_3}\)-type structures. Group \(16\) elements have \(6\) valence electrons and need \(2\) more. Group \(17\) elements have \(7\) valence electrons and need only \(1\) more. The hydride formula pattern therefore reflects the octet requirement across the non-metal side. The change is based on valence-shell structure, not on neutron count.
413. The table gives formula clues and their periodic meanings.
| Row | Formula clue | Periodic meaning |
|---|
| P | \(\mathrm{MH}\) for an alkali metal hydride | Metal commonly forms \(M^+\) |
| Q | \(\mathrm{H_2E}\) for a group \(16\) hydride | Element commonly has simple valency \(2\) |
| R | \(\mathrm{EO_3}\) with oxygen as \(-2\) | Element is in oxidation state \(+6\) |
| S | \(\mathrm{HX}\) for a halogen hydride | Halogen has simple valency \(4\) |
The row that needs correction is
ⓐ. Row S
ⓑ. Row P
ⓒ. Row Q
ⓓ. Row R
Correct Answer: Row S
Explanation: Row S needs correction because a halogen hydride \(\mathrm{HX}\) shows that the halogen has simple valency \(1\), not \(4\). Row P is correct because an alkali metal hydride formula \(\mathrm{MH}\) is consistent with \(M^+\) and \(\mathrm{H^-}\). Row Q correctly links \(\mathrm{H_2E}\) with valency \(2\) for group \(16\)-type elements. Row R is also correct because in \(\mathrm{EO_3}\), three oxygen atoms contribute \(-6\), so \(E\) is \(+6\). Formula interpretation must use charge balance or valency logic, not only the number of symbols in the formula.
414. Read the short classification case.
Two unknown elements \(A\) and \(B\) form hydrides of the type \(\mathrm{AH_3}\) and \(\mathrm{BH_3}\). Their highest simple oxides are also of the same formula type. Their atomic masses are different, but their compounds show related formula patterns.
The strongest classification inference is that \(A\) and \(B\) probably
ⓐ. are the same isotope
ⓑ. must have equal atomic numbers
ⓒ. belong to the same chemical family or group
ⓓ. are unrelated because their atomic masses differ
Correct Answer: belong to the same chemical family or group
Explanation: Similar hydride and oxide formulae suggest similar combining capacity and related chemical behaviour. Elements in the same group often form compounds of similar formula types because their valence-shell configurations are similar. Different atomic masses do not prevent elements from belonging to the same group. Isotopes would have the same atomic number, but the passage describes two elements, not atoms of one element. The formula pattern is a periodic-family clue rather than proof of identity.
415. Diagonal relationship refers to similarity between certain elements that are
ⓐ. diagonally adjacent in the periodic table, especially between period \(2\) and period \(3\)
ⓑ. in the same isotope series
ⓒ. always in the same vertical group only
ⓓ. arranged alphabetically beside each other
Correct Answer: diagonally adjacent in the periodic table, especially between period \(2\) and period \(3\)
Explanation: Diagonal relationship is observed between some elements of period \(2\) and diagonally adjacent elements of period \(3\). The similarity is not the same as ordinary group similarity, because the elements are not in the same vertical column. Examples include \(\mathrm{Li}\)-\(\mathrm{Mg}\), \(\mathrm{Be}\)-\(\mathrm{Al}\), and \(\mathrm{B}\)-\(\mathrm{Si}\). The relationship arises from a balance of size, charge density, electronegativity, and polarizing power. It is a positional and property-based similarity, not an alphabetical connection.
416. The pair that is a standard example of diagonal relationship is
ⓐ. \(\mathrm{Na}\) and \(\mathrm{K}\)
ⓑ. \(\mathrm{Li}\) and \(\mathrm{Mg}\)
ⓒ. \(\mathrm{F}\) and \(\mathrm{Cl}\)
ⓓ. \(\mathrm{Ne}\) and \(\mathrm{Ar}\)
Correct Answer: \(\mathrm{Li}\) and \(\mathrm{Mg}\)
Explanation: \(\mathrm{Li}\) and \(\mathrm{Mg}\) are diagonally related because lithium is in period \(2\), group \(1\), while magnesium is in period \(3\), group \(2\). They show some similarities that are not expected from simple vertical group comparison alone. \(\mathrm{Na}\) and \(\mathrm{K}\), \(\mathrm{F}\) and \(\mathrm{Cl}\), and \(\mathrm{Ne}\) and \(\mathrm{Ar}\) are same-group comparisons rather than diagonal pairs. Diagonal relationship is most notable among the lighter elements where size and charge-density effects are strong.
417. Use the arrangement described below: \(\mathrm{Li}\) is placed in period \(2\), group \(1\); \(\mathrm{Be}\) in period \(2\), group \(2\); \(\mathrm{Mg}\) in period \(3\), group \(2\); and \(\mathrm{Al}\) in period \(3\), group \(13\). The diagonal pairs in this arrangement are
ⓐ. \(\mathrm{Li}\)-\(\mathrm{Mg}\) and \(\mathrm{Be}\)-\(\mathrm{Al}\)
ⓑ. \(\mathrm{Li}\)-\(\mathrm{Be}\) and \(\mathrm{Mg}\)-\(\mathrm{Al}\)
ⓒ. \(\mathrm{Li}\)-\(\mathrm{Al}\) and \(\mathrm{Be}\)-\(\mathrm{Mg}\)
ⓓ. \(\mathrm{Mg}\)-\(\mathrm{Be}\) only
Correct Answer: \(\mathrm{Li}\)-\(\mathrm{Mg}\) and \(\mathrm{Be}\)-\(\mathrm{Al}\)
Explanation: A diagonal pair is formed when the second element is one period lower and one group to the right in the relevant early-table region. From \(\mathrm{Li}\), moving diagonally down-right gives \(\mathrm{Mg}\). From \(\mathrm{Be}\), moving diagonally down-right gives \(\mathrm{Al}\). Horizontal pairs such as \(\mathrm{Li}\)-\(\mathrm{Be}\) are period neighbours, not diagonal relationships. The arrangement must be read by both row and column movement.
418. The similarity in diagonal pairs such as \(\mathrm{Li}\)-\(\mathrm{Mg}\) is commonly explained by comparable
ⓐ. identical atomic number
ⓑ. charge/radius ratio and polarizing power
ⓒ. identical number of occupied shells
ⓓ. complete noble-gas configuration in the neutral atoms
Correct Answer: charge/radius ratio and polarizing power
Explanation: Diagonal relationship is not based on identical atomic number or identical shell number. It is often explained by a balance between increasing size down a group and increasing charge or different effective attraction across a period. This can produce comparable charge/radius ratio and polarizing power in certain diagonal pairs. Similar polarizing power can make some compounds show related behaviour. The similarity is partial and specific, not a claim that the two elements are chemically identical.
419. A claim says, “Diagonal relationship is just the same as group similarity.” The best correction is that diagonal relationship
ⓐ. compares only isotopes of the same element
ⓑ. requires both elements to have the same group number
ⓒ. compares certain diagonally adjacent elements whose properties become similar due to balancing periodic trends
ⓓ. is based only on atomic mass equality
Correct Answer: compares certain diagonally adjacent elements whose properties become similar due to balancing periodic trends
Explanation: Same-group similarity comes from similar valence-shell configurations in a vertical column. Diagonal relationship is different because the elements are in adjacent periods and neighbouring groups. In early periodic-table regions, changes in size, electronegativity, and charge density can balance in a way that creates partial similarity. Examples include \(\mathrm{Li}\)-\(\mathrm{Mg}\) and \(\mathrm{Be}\)-\(\mathrm{Al}\). The relationship should be treated as a special comparison, not as a replacement for group trends.
420. Assertion: \(\mathrm{Be}\) and \(\mathrm{Al}\) show diagonal relationship.
Reason: They have exactly the same electronic configuration and the same atomic number.
ⓐ. Both Assertion and Reason are true, and Reason explains Assertion
ⓑ. Assertion is true, but Reason is false
ⓒ. Both Assertion and Reason are true, but Reason does not explain Assertion
ⓓ. Assertion is false, but Reason is true
Correct Answer: Assertion is true, but Reason is false
Explanation: \(\mathrm{Be}\) and \(\mathrm{Al}\) are a standard diagonal pair in the periodic table. Their similarity is linked with comparable charge density, polarizing power, and some related compound behaviour. The Reason is false because \(\mathrm{Be}\) and \(\mathrm{Al}\) do not have the same atomic number or the same electronic configuration. Diagonal relationship never requires element identity. It is a partial similarity produced by balancing trends across a period and down a group.
