As ice is the solid form of water and it has more hydrogen bonds than water because its oxygen atoms are precisely tetrahedrally positioned and each oxygen is hydrogen bonded by four neighbouring oxygen atoms.
This changes the density of ice by expanding it, but does it change the strength of the hydrogen bonds of ice? Or more precisely, which one has strong hydrogen bonds, water or ice?
The quick answer is that ice has stronger hydrogen bonds than liquid water on average.$\require{mediawiki-texvc}%for angstrom unit$
As has already been brought up, this question is not easy to answer comprehensively as it may very well depend on temperature and pressure in ways that are not immediately obvious. That being said, I will answer this question for the most obvious case which is looking at hydrogen bond strength when water is just barely a liquid (just above $\pu{0~^\circ C}$) and when water is just barely frozen (just below $\pu{0~^\circ C}$). Also, as you may or may not know, ice freezes into many different forms which all have different lattice arrangements. I cannot speak to unusual forms of ice such as the one with a cubic lattice, but what I say is immediately applicable to forms like proton-ordered hexagonal ice (I think this is ice IX). Regular ice is called ice Ih.
It is worth looking at the very comprehensive reference [1], which discusses details of water from the dimer all the way to the many forms of ice that can form. The important points for us are only the following structural values, however:
In ice Ih, the average $\ce{O-H}$ bond length is $\pu{1.01 \AA}$ and the $\ce{HOH}$ bond angle is $109.5^\circ$. The trend that has been observed experimentally is that in going from the gas-phase water monomer, through the clusters, to the liquid, and finally to ice, the bond angle slowly increases to this tetrahedral value. Additionally, the average bond length in the liquid is shorter than in the solid, but you will have a hard time finding reliable reported values of this number from experiment because it is very difficult to determine the position of hydrogen atoms from scattering experiments in non-solid phases (this is mostly because of an ambiguity as to which oxygen atoms a hydrogen should be associated with in the scattering data). Nonetheless, there are reliable average $\ce{O-O}$ distances from scattering experiments.
In the liquid, the average $\ce{O-O}$ distance is $\pu{2.85 \AA}$ while in ice Ih it shrinks to $\pu{2.74 \AA}$. Intuitively, it is very hard to understand how this would happen if it is not simply due to increased hydrogen bond strength.
We can look at an even simpler case though and see that increased hydrogen bond strength causes the $\ce{O-O}$ distance to decrease in ice. This example is a comparison of the water dimer and water trimer. In the water dimer, the $\ce{O-O}$ distance is $\pu{2.976 \AA}$. In the water trimer, the $\ce{O-O}$ distance is about $\pu{2.82 \AA}$. This decrease in distance and the corresponding red shift in vibrational frequency are given physical meaning in the Ref. [2].
Ref. [2] gets at a very important point for us, one which Ref. [1] also uses to explain the shrinking in $\ce{O-O}$ distances in ice. This is because hydrogen bonding in water is highly cooperative. What do I mean by this? This is exactly the topic of Ref. [2]. The key idea is this: when you allow three water molecules to mutually hydrogen bond, as in the ring shape of the water trimer, the increase in attractive interaction strength is more than simply three times the interaction strength of a water dimer. By attractive interaction strength, I mean that if you subtract out the initial decrease in binding energy that comes from distorting the water monomers. Again, all of this is laid out in detail in Ref. [2], but we learn the general rule that many-body interactions in water are very important.
So, finally, I feel we have said enough to explain why the knowledge that a shrinking $\ce{O-O}$ distance and widening $\ce{HOH}$ angle tells us that ice has stronger hydrogen bonds than liquid water. In the first case, the decreasing $\ce{O-O}$ distance means that every pair of monomers experiences greater nuclear repulsion. This must be at least offset by a greater hydrogen bond strength in order for this geometry to be favorable. Part of the way this repulsion is offset is that as the system cools and forms a solid, the hydrogen bonds are allowed to become more ideal. This is why the bond angle widens. The tetrahedral arrangement of atoms means that the $\ce{OH--H}$ hydrogen-bond angle is much closer to 180 degrees on average than in the liquid where it frequently deviates fairly significantly.
There is one last point to make. It can be estimated both from accurate simulations and from experiment what is the average dipole moment of a water monomer in bulk liquid and in ice. It is found that the dipole increases going from the liquid to ice. This additionally tells us that the average charge on each hydrogen atom must have increased from in the liquid. We are guaranteed this because we know the bond angle increased, which should generally decrease the dipole moment (a linear configuration has no dipole). Instead, the bond angle increases, so we get more ideal hydrogen bonds, and the charge increases, so the hydrogen bonds would be stronger even if they were no more ideal than in the liquid!
Hopefully, this rather long-winded answer helps you understand how we can take the evidence at hand to arrive at the conclusion that ice has stronger hydrogen bonds than liquid water.
Now to the question of different forms of ice and different temperatures and pressures. I would think that pretty much across the board, every form of ice will follow the same trends we have seen here and hence have stronger hydrogen bonds. One strange case is cubic ice, which has $\ce{OHO}$ angles somewhere close to 90 degrees. This is very strange, but should lead to a large increase in the dipole of each molecule and probably an increase in positive charge on each hydrogen atom, so even then the hydrogen bonds should be stronger than in the liquid.
There may be temperatures and pressures where some of these arguments break down, but the $\ce{O-O}$ distance should smoothly decrease to that in the liquid and the angle should close to that in the liquid as the pressure increases, so one would expect the hydrogen bond strength to decrease to that in the liquid and never overshoot it.
References:
Ludwig, R. Water: From Clusters to the Bulk. Angew. Chem. Int. Ed. 2001, 40 (10), 1808-1827. DOI: 10.1002/1521-3773(20010518)40:10<1808::AID-ANIE1808>3.0.CO;2-1.
Xantheas, S. S. Cooperativity and hydrogen bonding network in water clusters. Chemical Physics 2000, 258 (2-3), 225-231. DOI: 10.1016/S0301-0104(00)00189-0.
TL;DR: it has been demonstrated both thermodynamically and computationally that the hydrogen bonding in hexagonal ("ordinary") ice is greater than in liquid water.
Ford and Falk [1] used spectroscopic data (IR absorption frequencies) associated with hydrogen bond strength (intermolecular potential energy). Namely, the relationship between the measured frequency shift $\Delta\nu$ of OH stretching vibration and correlation curves for the variety of frequencies allows us to plot the distribution of hydrogen bond energies in ice and water. Liquid water possesses a much broader variety of energies (only about $15\,\%$ molecules lie within $\pu{± 1 kcal mol^-1}$ range at maxima) in comparison to ice $(90\,\%).$ The authors partially attribute greater hydrogen bonding energy for ice to the heat of fusion $(\pu{1.435 kcal mol^-1}).$
Huš and Urbic [2] mention in their review (reference numbers have been adapted to the current references list) that
The HB strength in ice is greater than in liquid water [3–5]
and by performing own DFT calculations (6-31++G(df) and aug-cc-pVTZ basis sets) for the water cluster with the first and second solvation shells also backs this observation up. It was found that the addition of water molecules results in an additive change in bond strength, about $\pu{± 0.7 kcal mol^−1}$ per added molecule being a basis-agnostic value. As the solvation shell used in calculation expands, the average hydrogen energy increases, too (alongside the number of highly unlikely coordination environments with extremely high/low values):
When all possible hydrogen bonds are formed, the structure of the cluster resembles a highly ordered structure of ice. In that instance HB strength was calculated to be $\pu{7.84 kcal mol−1}$ (basis set 6-31++G(df)) or $\pu{6.66 kcal mol−1}$ (basis set aug-cc-pVTZ).
Authors also reference experimentally measured hydrogen bond strength of $\pu{3.6 ± 0.5 kcal mol^−1}$ using thermal conductivity of the vapor [6].
In liquid state of water the atoms will be having so much kinetic energy that they cannot form permanent bonds.
although bonds will be formed but they will break simultaneously until the temperature hence the kinetic energy is reduced and bonds become stable
Although the strength will be almost the same but due to constant breaking and joining the net interaction will be weaker than in ice
The hydrogen bonding in ice(hex) is about 3 kJ ˣ mol-1 stronger than liquid water. According to this paper http://www1.lsbu.ac.uk/water/water_hydrogen_bonding.html
this is a nice video i found on youtube :) https://www.youtube.com/watch?v=UukRgqzk-KE
Water has stronger hydrogen bonds than ice does.
Liquid water is denser than ice. Since water and ice are both made of H2O molecules, the fact that water is denser means the H2O molecules are closer together in water than they are in ice. This means the intermolecular forces attracting one H2O molecule to another must be stronger in water than they are in ice. Since hydrogen bonds are the primary intermolecular forces in H2O, the hydrogen bonds in liquid water are stronger than those in ice.
Remember that hydrogen bonds are an electrostatic attraction (an attraction between a positive charge and a negative charge). The strength of an electrostatic attraction depends on the magnitude of the charges and on the distance between the charges. Since the distance between molecules is greater in ice but the magnitude of the charge is the same, the strength of the attraction is lower.
