Hydrogen — NEET Chemistry MCQs with answers

Hydrogen is a Class 11 Chemistry chapter in the NEET (UG) syllabus. NEET720 has 122 reviewed practice questions on it, each with a quick answer and a step-by-step explanation. The 8 questions below are free and fixed, so you can bookmark this page; the full chapter, plus mistake tracking and spaced revision, is in the app.

Topics covered

Preparation of dihydrogen · Uses of dihydrogen · Dihydrogen · Position of Hydrogen · Dihydrogen Preparation · Water · Heavy Water · Dihydrogen Properties · Dihydrogen Uses · Position of hydrogen in periodic table · Types of hydrides · Hydrogen peroxide industrial preparation · Heavy water properties · Water hardness · Ionic hydride reaction with water · Ortho and para hydrogen · Hydrides classification · Hydrogen as a fuel / hydrogen economy · Hydrides — comparison of stability · Hydrogen bonding and water anomalies · Hydrogen — reducing property · Hydrogen bonding types · Hydrogen peroxide — structure · Hydrogen — industrial preparation (Bosch process) · Hydrogen · Hydrides · Properties of Water · Hydrogen Peroxide · Isotopes of Hydrogen · Hydrogen Economy · Hydrogen as a Fuel

8 free Hydrogen practice questions with answers

Choose an answer in your head before opening it. Each explanation says why the correct option is right and, where relevant, why the tempting wrong option is wrong.

1. Question 1 · easy · Position of Hydrogen

   Hydrogen is sometimes placed in group 1 (with alkali metals) and sometimes in group 17 (with halogens) of the periodic table, reflecting its unique dual character. The resemblance of hydrogen to alkali metals is primarily based on:

   • A.Both have a single electron in their outermost shell (ns1) and can lose this electron to form a +1 cation
   • B.Both hydrogen and alkali metals are typically liquids at room temperature
   • C.Both have exactly 7 electrons in their outermost shell
   • D.Both hydrogen and alkali metals are strongly electronegative, non-metallic elements
   Show answer and explanation

   Answer: A. Both have a single electron in their outermost shell (ns1) and can lose this electron to form a +1 cation

   Both hydrogen (1s1) and alkali metals (ns1) have a single electron in their outermost shell, and both can lose this electron to form a stable +1 cation, forming the basis for grouping hydrogen with the alkali metals.

   Hydrogen has the simplest possible electronic configuration, 1s1, with a single electron in its only shell. This parallels the outer electronic configuration of alkali metals (Li: [He]2s1, Na: [Ne]3s1, etc.), which also have exactly one electron in their outermost s orbital. Both hydrogen and alkali metals can lose this single outer electron to form a stable +1 cation (H+ and M+ respectively, though H+ is simply a bare proton, quite different in practice from a full alkali metal cation), and both can act as reducing agents in various reactions. This structural/electronic similarity is the basis for placing hydrogen above the alkali metals in some periodic table representations, despite hydrogen showing many properties quite distinct from typical metals (it is a gas, non-metallic in many of its reactions, and does not show the strong electropositive metallic character of true alkali metals).

   Common mistake: Confusing the electronic basis for hydrogen's alkali-metal-like behaviour with its halogen-like behaviour (which instead relates to needing one more electron to complete a duplet)

2. Question 2 · easy · Position of Hydrogen

   Hydrogen's resemblance to the halogens (group 17) is primarily based on the fact that:

   • A.Both hydrogen and halogens are one electron short of achieving a stable noble gas (duplet/octet) configuration, and both can gain an electron to form a -1 anion (hydride, H−, and halide, X−)
   • B.Both hydrogen and halogens exist as monatomic gases under all conditions
   • C.Both hydrogen and halogens have exactly 8 electrons in their valence shell
   • D.Both hydrogen and halogens are strongly metallic in character
   Show answer and explanation

   Answer: A. Both hydrogen and halogens are one electron short of achieving a stable noble gas (duplet/octet) configuration, and both can gain an electron to form a -1 anion (hydride, H−, and halide, X−)

   Both hydrogen (needing 1 more electron to reach the stable duplet of helium, 1s2) and halogens (needing 1 more electron to reach a stable octet) are one electron short of a noble gas configuration, and both readily gain this electron to form a stable -1 anion (hydride or halide).

   Hydrogen (1s1) requires just one additional electron to achieve the stable duplet electronic configuration of helium (1s2), the nearest noble gas. Halogens (ns2np5) similarly require just one additional electron to achieve a stable octet configuration matching the nearest noble gas. Both hydrogen and halogens can therefore gain one electron to form a stable, singly-charged anion: hydrogen forms the hydride ion (H−), and halogens form the corresponding halide ions (F−, Cl−, Br−, I−). Additionally, like halogens, hydrogen also exists as a diatomic molecule (H2) under standard conditions, held together by a covalent single bond, similar in general character to the diatomic halogen molecules (F2, Cl2, etc.), further supporting this comparison, which forms the basis for occasionally placing hydrogen above the halogens in some periodic table representations, despite hydrogen also showing the alkali-metal-like similarities discussed earlier.

   Common mistake: Confusing the specific electron count/configuration reasoning that underlies hydrogen's halogen-like resemblance with its alkali-metal-like resemblance

3. Question 3 · medium · Isotopes of Hydrogen

   Hydrogen has three naturally occurring/known isotopes: protium (1H, no neutrons), deuterium (2H or D, 1 neutron), and tritium (3H or T, 2 neutrons, radioactive). Reactions involving deuterium generally proceed somewhat SLOWER than the corresponding reactions of ordinary hydrogen (protium), a phenomenon called the kinetic isotope effect. This is primarily explained by:

   • A.The greater mass of deuterium results in a lower zero-point vibrational energy for bonds involving D compared to bonds involving H, requiring more energy (higher activation energy) to break the D-containing bond
   • B.Deuterium has a different number of protons than protium, giving it entirely different chemical properties
   • C.Deuterium reacts faster than protium in all chemical reactions, contrary to the stated kinetic isotope effect
   • D.The kinetic isotope effect has no established scientific explanation and is a purely empirical, unexplained curiosity
   Show answer and explanation

   Answer: A. The greater mass of deuterium results in a lower zero-point vibrational energy for bonds involving D compared to bonds involving H, requiring more energy (higher activation energy) to break the D-containing bond

   Because deuterium is heavier than protium, a bond to deuterium (like C-D) vibrates at a lower zero-point energy than the corresponding bond to protium (C-H); this means more energy must be supplied to break the C-D bond (reach the transition state) compared to breaking a C-H bond, resulting in a higher effective activation energy and hence slower reaction rate for deuterium-substituted reactions.

   According to quantum mechanics, even at absolute zero, a chemical bond retains some minimum vibrational energy, called the zero-point energy, which depends on the reduced mass of the vibrating atoms (heavier atoms/isotopes give a lower zero-point energy for the same bond, since the vibrational frequency is inversely related to the square root of the reduced mass). Since deuterium is approximately twice as heavy as protium, a bond involving deuterium (such as C-D) has a lower zero-point vibrational energy than the corresponding bond involving protium (C-H), meaning the C-D bond sits in a slightly deeper energy well and thus effectively has a somewhat higher bond dissociation energy. Consequently, reactions that involve breaking this particular bond in the rate-determining step require overcoming a slightly higher activation energy barrier when deuterium is present compared to when ordinary hydrogen is present, resulting in the observed kinetic isotope effect where deuterium-substituted reactions typically proceed measurably slower than the corresponding protium reactions, a phenomenon widely used by chemists to help elucidate reaction mechanisms (by determining whether a specific bond-breaking step is rate-determining).

   Common mistake: Assuming isotope effects arise from a difference in fundamental chemical identity or proton count rather than the correct mass-dependent zero-point energy explanation

4. Question 4 · medium · Dihydrogen Preparation

   Industrially, large quantities of dihydrogen are produced via the steam reforming of natural gas (methane), involving the reaction: CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g) (at high temperature, ~1270 K, with a nickel catalyst), followed by the water-gas shift reaction: CO(g) + H2O(g) ⇌ CO2(g) + H2(g) (at lower temperature, with an iron oxide or copper catalyst). The purpose of this second (water-gas shift) reaction is to:

   • A.Convert the remaining carbon monoxide byproduct into additional hydrogen gas (plus easily separable CO2), maximizing overall hydrogen yield from the process
   • B.Remove all hydrogen gas produced in the first step, converting it back into methane
   • C.Introduce additional carbon monoxide into the product stream for use in a separate industrial process
   • D.Convert methane directly into carbon dioxide without producing any hydrogen at all
   Show answer and explanation

   Answer: A. Convert the remaining carbon monoxide byproduct into additional hydrogen gas (plus easily separable CO2), maximizing overall hydrogen yield from the process

   The water-gas shift reaction reacts the CO byproduct from steam reforming with additional steam to produce more H2 gas (plus CO2, which can be easily removed by scrubbing), significantly increasing the overall hydrogen yield obtainable from a given quantity of methane feedstock.

   The industrial production of hydrogen via steam reforming occurs in two sequential stages. First, methane reacts with steam over a nickel catalyst at high temperature to produce a mixture of carbon monoxide and hydrogen (called synthesis gas or syngas): CH4 + H2O ⇌ CO + 3H2. While this first step already produces significant hydrogen, the CO byproduct represents 'wasted' potential hydrogen yield if not further processed. The second stage, the water-gas shift reaction, reacts this CO with additional steam over a suitable catalyst (typically iron oxide-chromium oxide at higher temperature, or copper-zinc oxide at lower temperature for a more complete conversion) to convert it into carbon dioxide and additional hydrogen gas: CO + H2O ⇌ CO2 + H2. This second reaction significantly increases the overall hydrogen yield obtainable from the original methane feedstock, and the resulting CO2 byproduct can be relatively easily separated from the hydrogen product stream (e.g., by scrubbing with a suitable absorbent), leaving a purified hydrogen product.

   Common mistake: Not recognizing that the water-gas shift reaction is specifically designed to extract ADDITIONAL hydrogen from the CO byproduct, mistakenly thinking it consumes or reverses hydrogen production

5. Question 5 · medium · Hydrides

   Hydrides are broadly classified into ionic (saline), covalent (molecular), and metallic (interstitial) types. NaH, CH4, and PdH0.6 (an example of a metal hydride) each represent one of these categories respectively. This classification is based primarily on:

   • A.The electronegativity difference between hydrogen and the element it combines with, and whether hydrogen occupies interstitial spaces in a metal lattice
   • B.The physical state (solid, liquid, or gas) of the hydride at room temperature, unrelated to bonding character
   • C.All three types of hydrides (ionic, covalent, metallic) have identical bonding character, differing only in their names
   • D.The classification depends only on which specific hydrogen isotope (protium, deuterium, tritium) is present in the hydride
   Show answer and explanation

   Answer: A. The electronegativity difference between hydrogen and the element it combines with, and whether hydrogen occupies interstitial spaces in a metal lattice

   Ionic (saline) hydrides form when hydrogen combines with a highly electropositive metal (like Na), gaining an electron to form H− in an ionic lattice; covalent hydrides form when hydrogen combines with a similarly electronegative nonmetal (like C) via electron sharing; metallic (interstitial) hydrides form when small hydrogen atoms occupy interstitial voids within a transition metal's crystal lattice.

   Ionic (saline) hydrides form when hydrogen reacts with highly electropositive s-block metals (Group 1 and heavier Group 2 metals), where the large electronegativity difference favours essentially complete electron transfer from the metal to hydrogen, forming the hydride ion H− within an ionic crystal lattice (e.g., NaH, CaH2). Covalent (molecular) hydrides form when hydrogen combines with p-block nonmetals (and some metalloids) of comparable electronegativity, sharing electrons to form discrete covalent molecules (e.g., CH4, NH3, H2O, HCl). Metallic (interstitial) hydrides form specifically with many transition metals (particularly those from groups 3-12, especially d-block metals like palladium, titanium, and various lanthanoids), where small hydrogen atoms occupy the interstitial voids within the metal's existing crystal lattice (analogous to the interstitial compounds discussed with carbon/nitrogen earlier), often giving non-stoichiometric compositions (like PdH0.6, reflecting the specific number of interstitial sites actually occupied) while the metal generally retains its metallic conductivity and lustre. This three-way classification directly parallels the electronegativity-based reasoning used elsewhere in inorganic chemistry (ionic bonding for large electronegativity difference, covalent bonding for comparable electronegativity), with the metallic/interstitial category representing hydrogen's distinctive additional behaviour specifically with transition metal lattices.

   Common mistake: Attributing hydride classification to physical state or hydrogen isotope rather than the correct bonding-character-based reasoning

6. Question 6 · medium · Hydrides

   Ionic hydrides like NaH react vigorously with water, releasing hydrogen gas: NaH + H2O → NaOH + H2. In this reaction, the hydride ion (H−) acts specifically as:

   • A.A strong Bronsted base, accepting a proton from water (which acts as the Bronsted acid)
   • B.A Bronsted acid, donating a proton to water
   • C.An oxidizing agent, being reduced further in this reaction
   • D.A spectator ion with no active chemical role in this reaction
   Show answer and explanation

   Answer: A. A strong Bronsted base, accepting a proton from water (which acts as the Bronsted acid)

   The hydride ion (H−), possessing a complete electron pair and a full negative charge, acts as an extremely strong Bronsted base, readily accepting a proton from water (H2O acting as the Bronsted acid) to form H2 gas and hydroxide ion.

   The hydride ion, H−, consists of a hydrogen nucleus (a single proton) plus two electrons (isoelectronic with helium), giving it a complete octet-analogous (duplet) stable electron configuration and an overall -1 charge. In the reaction NaH + H2O → NaOH + H2, the hydride ion acts as an extremely strong Bronsted base (proton acceptor): it accepts a proton (H+) from a water molecule, forming a new H-H bond (giving H2 gas), while the water molecule, having donated a proton, becomes hydroxide ion (OH−), combining with the spectator Na+ ion to give NaOH. This vigorous, essentially irreversible reaction (releasing flammable hydrogen gas and generating significant heat) reflects the hydride ion's status as one of the strongest known Bronsted bases, readily reacting even with weak proton sources like water.

   Common mistake: Reversing the acid-base roles, incorrectly describing the hydride ion as a proton donor rather than a proton acceptor

7. Question 7 · easy · Water

   Temporary hardness of water, caused by dissolved bicarbonates of Ca2+ and Mg2+, can be removed simply by boiling, while permanent hardness, caused by dissolved chlorides and sulfates of Ca2+/Mg2+, cannot be removed by boiling alone. This difference is because:

   • A.Boiling decomposes soluble bicarbonates into insoluble carbonates (which precipitate out), but chlorides and sulfates of Ca2+/Mg2+ remain soluble and stable even at boiling temperature
   • B.Boiling has no chemical effect on bicarbonates at all; temporary hardness is removed by physical evaporation of water alone
   • C.Chlorides and sulfates of calcium/magnesium decompose more easily on boiling than bicarbonates do
   • D.Temporary and permanent hardness are chemically identical, with no real distinction between them
   Show answer and explanation

   Answer: A. Boiling decomposes soluble bicarbonates into insoluble carbonates (which precipitate out), but chlorides and sulfates of Ca2+/Mg2+ remain soluble and stable even at boiling temperature

   Boiling water containing dissolved Ca(HCO3)2/Mg(HCO3)2 causes thermal decomposition, releasing CO2 gas and precipitating insoluble CaCO3/MgCO3 (removing the hardness-causing ions from solution), while chloride and sulfate salts of Ca2+/Mg2+ remain fully soluble and stable at boiling temperature, so this method cannot remove permanent hardness.

   Temporary hardness is caused by dissolved calcium and magnesium bicarbonates, Ca(HCO3)2 and Mg(HCO3)2, which are only moderately soluble and thermally unstable. Upon boiling, these bicarbonates decompose: Ca(HCO3)2 → CaCO3(insoluble precipitate) + H2O + CO2(gas), and similarly for magnesium bicarbonate, effectively removing the Ca2+/Mg2+ ions from solution as an insoluble precipitate that can be filtered off, thereby removing the hardness. Permanent hardness, in contrast, is caused by dissolved calcium and magnesium chlorides and sulfates (CaCl2, MgCl2, CaSO4, MgSO4), which are considerably more soluble and thermally stable salts that do not decompose or precipitate simply upon boiling; these ions remain dissolved in solution regardless of boiling, requiring alternative methods (such as the addition of washing soda/Na2CO3, ion-exchange resins, or the Clark's process using slaked lime) to remove this type of hardness.

   Common mistake: Assuming boiling is merely a physical process rather than recognizing the specific chemical decomposition of bicarbonate salts responsible for temporary hardness removal

8. Question 8 · medium · Water

   Clark's process removes temporary hardness by adding a calculated amount of slaked lime, Ca(OH)2, to hard water: Ca(HCO3)2 + Ca(OH)2 → 2CaCO3↓ + 2H2O. This method works effectively because:

   • A.The hydroxide ion reacts with the bicarbonate ion to precipitate calcium carbonate, and if the exact calculated (not excess) amount of lime is used, the calcium already present in the original hardness plus the calcium introduced from the lime are both effectively converted to and removed as insoluble CaCO3 precipitate
   • B.Adding lime always increases the hardness of water, since more calcium is being added to the water
   • C.Clark's process works only on permanent hardness, not on temporary hardness
   • D.The reaction produces soluble calcium bicarbonate as the final product, leaving hardness unchanged
   Show answer and explanation

   Answer: A. The hydroxide ion reacts with the bicarbonate ion to precipitate calcium carbonate, and if the exact calculated (not excess) amount of lime is used, the calcium already present in the original hardness plus the calcium introduced from the lime are both effectively converted to and removed as insoluble CaCO3 precipitate

   Adding a precisely calculated amount of slaked lime reacts with the dissolved calcium bicarbonate to precipitate insoluble calcium carbonate, effectively removing both the originally-present hardness-causing calcium and the calcium introduced by the lime itself from solution, provided the amount of lime added is carefully calculated (not in excess) to match the bicarbonate content.

   In Clark's process, a carefully calculated amount of slaked lime, Ca(OH)2, is added to hard water containing dissolved calcium (or magnesium) bicarbonate. The hydroxide ions from the lime react with the bicarbonate ions, converting them into insoluble carbonate precipitate: Ca(HCO3)2 + Ca(OH)2 → 2CaCO3↓ + 2H2O. Notably, this reaction consumes ONE mole of the original hardness-causing calcium bicarbonate and ONE mole of the added lime calcium hydroxide to produce TWO moles of insoluble calcium carbonate precipitate — meaning both the calcium originally present as hardness AND the calcium introduced via the lime reagent are together removed from solution as the precipitate, resulting in a net REDUCTION of total dissolved calcium (hardness) in the water, provided the amount of lime added is precisely calculated to match the bicarbonate content present (adding excess lime beyond this calculated amount would instead reintroduce excess dissolved Ca(OH)2, potentially increasing hardness again, which is why precise calculation/dosing is essential for this method to work effectively).

   Common mistake: Assuming adding a calcium-containing reagent must necessarily increase water hardness, without recognizing the precipitation mechanism that removes both sources of calcium

Questions about Hydrogen for NEET

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Questions are original NEET720 compositions reviewed for correctness, syllabus fit and option quality. Counts update as the bank grows (122 active practice questions in this chapter today).